āωāĻ¤ā§āϤāϰ āϭ⧁āϞ āĻŽāύ⧇ āĻšāĻ˛ā§‡Â  āύāĻŋāĻšā§‡ āĻŽāĻ¨ā§āϤāĻŦā§āϝ āĻ•āϰ⧁āύ

 

āĻĒāĻĻāĻžāĻ°ā§āĻĨāĻŦāĻŋāĻœā§āĻžāĻžāύ

 

 

ā§§. āĻ¸ā§āĻĨāĻŋāϰ āĻ…āĻŦāĻ¸ā§āĻĨāĻžā§Ÿ āĻĨāĻžāĻ•āĻž āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁ āĻŦāĻŋāĻ¸ā§āĻĢā§‹āϰāϪ⧇āϰ āĻĢāϞ⧇ M1 āĻāĻŦāĻ‚ M2 āĻ­āϰ⧇āϰ āĻĻ⧁āϟāĻŋ āĻ–āĻŖā§āĻĄā§‡ āĻŦāĻŋāĻ­āĻ•ā§āϤ āĻšā§Ÿ āĻāĻŦāĻ‚ āĻ–āĻŖā§āĻĄ āĻĻ⧁āϟāĻŋ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻāĻŋāϕ⧇ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡ V1 āĻāĻŦāĻ‚ V2 āĻŦ⧇āĻ— āĻĒā§āϰāĻžāĻĒā§āϤ āĻšā§ŸāĨ¤ V1 āĻāĻŦāĻ‚ V2 āĻāϰ āĻ…āύ⧁āĻĒāĻžāϤ āĻ•āϤ?

 

āĻ•. M1 / M2

āĻ–. M2 / M1

āĻ—. (M1 / M2)1/2

āϘ. (M1 / M2)1/2

 

āωāĻ¤ā§āϤāϰ :  āĻ–. M2 / M1

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

                 

āĻāĻ–āĻžāύ⧇, M1V1 = M2V2, ∴ V2 / V1 = M2 / M1

 

⧍. āĻāĻ•āϟāĻŋ āϘāϰ⧇āϰ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻā§â€™āĻĻā§‡ā§ŸāĻžāϞ⧇āϰ āĻŽāĻ§ā§āϝāĻŦāĻ°ā§āϤ⧀ āĻĻā§‚āϰāĻ¤ā§āĻŦ 4 m, āĻāĻ•āϟāĻŋ āĻĻā§‡ā§ŸāĻžāϞ⧇ āĻāĻ•āϟāĻŋ āĻ…āĻŦāϤāϞ āĻĻāĻ°ā§āĻĒāĻŖ āϞāĻžāĻ—āĻžāύ⧇āĻž āφāϛ⧇āĨ¤ āĻĻāĻ°ā§āĻĒāĻŖ āĻšāϤ⧇ 2.5 m āĻĻā§‚āϰ⧇ āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁ āϰāĻžāĻ–āϞ⧇ āϤāĻžāϰ āĻĒā§āϰāϤāĻŋāĻŦāĻŋāĻŽā§āĻŦ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻā§‡ā§ŸāĻžāϞ⧇ āĻ—āĻ āĻŋāϤ āĻšā§ŸāĨ¤ āĻĻāĻ°ā§āĻĒāϪ⧇āϰ āĻĢā§‹āĻ•āĻžāϏ āĻĻā§‚āϰāĻ¤ā§āĻŦ āĻ•āϤ?

 

āĻ•. 2.5 m

āĻ–. 1.54 m

āĻ—. 1.44 m

āϘ. 2.25 m

 

āωāĻ¤ā§āϤāϰ :  āĻ–. 1.54 m

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}, \therefore \mathrm{f}=\frac{u v}{u+v}=\frac{4 x 2.5}{4+2.5}=1.54$

 

ā§Š. āĻāĻ•āϟāĻŋ āϤ⧇āϜāĻ¸ā§āĻ•ā§āϰāĻŋ⧟ āĻĒāĻĻāĻžāĻ°ā§āĻĨ⧇āϰ āĻ…āĻ°ā§āϧāĻžā§Ÿā§ 1600 āĻŦāĻ›āϰāĨ¤ āĻ•āϤ āϏāĻŽā§Ÿ āĻĒāϰ⧇ āϤ⧇āϜāĻ¸ā§āĻ•ā§āϰāĻŋâ€Œā§Ÿ āĻĒāĻĻāĻžāĻ°ā§āĻĨ⧇āϰ 15/16 āĻ…āĻ‚āĻļ āĻ•ā§āώ⧟āĻĒā§āϰāĻžāĻĒā§āϤ āĻšāĻŦ⧇?

 

āĻ•. 1500 years

āĻ–. 4800 years

āĻ—. 6400 years

āϘ. 9600 years

 

āωāĻ¤ā§āϤāϰ :  āĻ—. 6400 years

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$\mathrm{t}_{1 / 2}=1600$ āĻŦāĻ›āϰ, 

$\therefore \lambda=\frac{0.693}{1600}$

āĻāĻ–āύ, $\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda \mathrm{t}}$ ,

āĻāĻ–āĻžāύ⧇ $\mathrm{N}=(1 / 16) \times \mathrm{N}_{0}, \therefore \mathrm{t}=-\frac{\ln \frac{1}{16}}{(0.693 / 1600)}=6400$ āĻŦāĻ›āϰ

 

ā§Ē. Higgs āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž āĻāĻ• āϧāϰāύ⧇āϰ -

 

āĻ•. āĻ­āϰ āϤ⧈āϰāĻŋāϰ āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž

āĻ–. āĻļāĻ•ā§āϤāĻŋ āϤ⧈āϰāĻŋāϰ āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž

āĻ—. āχāϞ⧇āĻ•āĻŸā§āϰāύ āϤ⧈āϰāĻŋāϰ āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž

āϘ. āĻŦāϞ āϤ⧈āϰāĻŋāϰ āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž

 

āωāĻ¤ā§āϤāϰ :  āĻ•. āĻ­āϰ āϤ⧈āϰāĻŋāϰ āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž

 

āϏ⧂āĻ¤ā§āϰ : Wikipedia

 

ā§Ģ. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āϭ⧌āϤ āĻĒā§āϰāĻ•ā§āϰāĻŋ⧟āĻž āĻļāĻŦā§āĻĻ āϤāϰāĻ™ā§āĻ— āĻĻā§āĻŦāĻžāϰāĻž āĻĒā§āϰāĻĻāĻ°ā§āĻļāĻŋāϤ āĻšā§Ÿ āύāĻž?

 

āĻ•. āĻĒā§āϰāϤāĻŋāϏāϰāĻŖ

āĻ–. āĻŦā§āϝāϤāĻŋāϚāĻžāϰ

āĻ—. āϏāĻŽāĻŦāĻ°ā§āϤāύ

āϘ. āĻ…āĻĒāĻŦāĻ°ā§āϤāύ

 

āωāĻ¤ā§āϤāϰ : āĻ—. āϏāĻŽāĻŦāĻ°ā§āϤāύ

 

āϏ⧂āĻ¤ā§āϰ : Wikipedia

 

ā§Ŧ. āĻĒ⧃āĻĨāĻŋāĻŦā§€āĻĒ⧃āĻˇā§āϠ⧇ āĻŽāĻšāĻžāĻ•āĻ°ā§āĻˇā§€ā§Ÿ āĻĒā§āϰāĻžāĻŦāĻ˛ā§āϝ g, āĻ•āĻžāĻ˛ā§āĻĒāύāĻŋāĻ• āĻāĻ•āϟāĻŋ āĻ—ā§āϰāĻšā§‡āϰ āϘāύāĻ¤ā§āĻŦ āϝāĻĻāĻŋ āĻĒ⧃āĻĨāĻŋāĻŦā§€āϰ āϘāύāĻ¤ā§āĻŦ⧇āϰ āϏāĻŽāĻžāύ āĻšā§Ÿ āĻāĻŦāĻ‚ āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ āϝāĻĻāĻŋ āĻĻā§āĻŦāĻŋāĻĒ⧁āĻŖ āĻšā§Ÿ āϤāĻŦ⧇ āĻāχ āĻ—ā§āϰāĻšā§‡āϰ āĻĒ⧃āĻˇā§āϠ⧇ āĻŽāĻšāĻžāĻ•āĻ°ā§āĻˇā§€ā§Ÿ āĻ•ā§āώ⧇āĻ¤ā§āϰ āĻĒā§āϰāĻžāĻŦāĻ˛ā§āϝ āĻ•āϤ?

 

āĻ•. g

āĻ–. 2g

āĻ—. 4g

āϘ. 8g

 

āωāĻ¤ā§āϤāϰ :  2g

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$g_{\text {earth }}=\frac{G M}{R^{2}}=\frac{G \rho\left(4 \pi R^{3}\right)}{R^{2}}=G \rho 4 \pi R, \therefore g_{\text {planet }}=G \rho 4 \pi(2 R)=2 \times g_{\text {earth }}$

 

ā§­. āĻāĻ•āϟāĻŋ āĻŦ⧈āĻĻā§āϝ⧁āϤāĻŋāĻ• āχāĻ¸ā§āĻ¤ā§āϰāĻŋāϤ⧇ 220 V āĻāĻŦāĻ‚ 1200 W āϞ⧇āĻ–āĻž āφāϛ⧇āĨ¤ āϝāĻĻāĻŋ āĻĒā§āϰāϤāĻŋ āχāωāύāĻŋāϟ āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻļāĻ•ā§āϤāĻŋāϰ āĻŽā§‚āĻ˛ā§āϝ 1.00 āϟāĻžāĻ•āĻž āĻšā§Ÿ, āϤāĻžāĻšāϞ⧇ āχāĻ¸ā§āĻ¤ā§āϰāĻŋāϟāĻŋ 2 āϘāĻŖā§āϟāĻž āϚāĻžāϞāĻžāϞ⧇ āĻ•āϤ āĻ–āϰāϚ āĻĒ⧜āĻŦ⧇?

 

āĻ•. 3 āϟāĻžāĻ•āĻž

āĻ–. 2.6 āϟāĻžāĻ•āĻž

āĻ—. 2.3 āϟāĻžāĻ•āĻž

āϘ. 2.4 āϟāĻžāĻ•āĻž

 

āωāĻ¤ā§āϤāϰ :  āϘ. 2.4 āϟāĻžāĻ•āĻž

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

P = 1200 W, t = 2 hour, ∴ āĻ–āϰāϚ = (Pt/1000) x 1.00 = 2.4 āϟāĻžāĻ•āĻž

 

ā§Ž. āύāĻŋāĻšā§‡āϰ āϞ⧇āĻ–āϚāĻŋāĻ¤ā§āϰ⧇ 50 s āϏāĻŽā§ŸāĻ•āĻžāϞ⧇ āĻāĻ•āϟāĻŋ āĻ—āĻžā§œāĻŋāϰ āĻŦ⧇āϗ⧇āϰ āĻĒāϰāĻŋāĻŦāĻ°ā§āϤāύ āĻĻ⧇āĻ–āĻžāύ⧋ āĻšā§Ÿā§‡āϛ⧇āĨ¤ āĻāχ āϏāĻŽā§ŸāĻ•āĻžāϞ⧇ āĻ—āĻžā§œāĻŋāϟāĻŋ āĻ•āϤ āĻĻā§‚āϰāĻ¤ā§āĻŦ āĻ…āϤāĻŋāĻ•ā§āϰāĻŽ āĻ•āϰ⧇āϛ⧇?

 

āĻ•. 500 m

āĻ–. 400 m

āĻ—. 350 m

āϘ. 300 m

 

āωāĻ¤ā§āϤāϰ : āĻ—. 350 m

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

āĻĒā§āϰāĻĨāĻŽ ⧍ā§Ļ āϏ⧇āϕ⧇āĻ¨ā§āĻĄā§‡ ā§§ā§Ļā§Ļ āĻ…āϤāĻŋāĻ•ā§āϰāĻŽ āĻ•āϰ⧇āĨ¤Â Â $\mathrm{S}=\frac{(0+v)}{2} x t=5 \times 20=100$

āĻĒāϰāĻŦāĻ°ā§āϤ⧀ ⧍ā§Ļ āϏ⧇āϕ⧇āĻ¨ā§āĻĄā§‡ s = vt = 200

āĻļ⧇āώ ā§§ā§Ļ āϏ⧇āϕ⧇āĻ¨ā§āĻĄā§‡ $\mathrm{S}=\frac{(0+v)}{2} x t=5 \times 20=100$, ∴ āĻŽā§‹āϟ = 350 m

 

⧝. 0°C āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžāϰ 2.1 kg āĻŦāϰāĻĢ 40°C āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžāϰ 5.9 kg āĻĒāĻžāύāĻŋāϰ āϏāĻžāĻĨ⧇ āĻŽāĻŋāĻļā§āϰāĻŋāϤ āĻ•āϰāĻž āĻšāϞ⧋āĨ¤ āĻŽāĻŋāĻļā§āϰāϪ⧇āϰ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻž āĻ•āϤ āĻšāĻŦ⧇? āĻĒāĻžāύāĻŋāϰ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āϤāĻžāĻĒ = 4.2x103 J Kg-1 K-1, āĻŦāϰāĻĢ āĻ—āϞāύ⧇āϰ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āϏ⧁āĻĒā§āϤāϤāĻžāĻĒ = 3.36x105 J Kg-1 K-1

 

āĻ•. 7.5°C

āĻ–. 9.5°C

āĻ—. 10.5°C

āϘ. 8.5°C

 

āωāĻ¤ā§āϤāϰ :  āϘ. 8.5°C

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$2.21 \times 3.36 \times 10^{5}+2.21 \times 4.2 \times 10^{3} \times \theta=5.9 \times 4.2 \times 10^{3} \times(40-\theta)$

āϏāĻŽāĻžāϧāĻžāύ āĻ•āϰ⧇, θ = 8.5°C

 

ā§§ā§Ļ. āĻĒāĻžāύāĻŋ āϏāĻžāĻĒ⧇āĻ•ā§āώ⧇ āĻ•āĻžāρāĻšā§‡āϰ āĻĒā§āϰāϤāĻŋāϏāĻžāϰāĻ™ā§āĻ• 9/8. āĻŦāĻžā§Ÿā§ āϏāĻžāĻĒ⧇āĻ•ā§āώ⧇ āĻ•āĻžāρāĻšā§‡āϰ āĻĒā§āϰāϤāĻŋāϏāϰāĻžāĻ™ā§āĻ• 3/2, āĻŦāĻžā§Ÿā§ āϏāĻžāĻĒ⧇āĻ•ā§āώ⧇ āĻĒāĻžāύāĻŋāϰ āĻĒā§āϰāϤāĻŋāϏāĻžāϰāĻ™ā§āĻ• āĻ•āϤ?

 

āĻ•. 2/3

āĻ–. 4/5

āĻ—. 4/3

āϘ. 3/4

 

āωāĻ¤ā§āϤāϰ :  āĻ—. 4/3

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$w \mu_{g}=a \mu_{g} / a \mu_{W}, \therefore a \mu_{W}=a \mu_{g} / w \mu_{g}=\frac{\frac{3}{2}}{\frac{9}{8}}=\frac{3 x 8}{2 x 9}=\frac{4}{3}$

 

ā§§ā§§. āϏāϰāϞ āĻ›āĻ¨ā§āĻĻāĻŋāϤ āĻ—āϤāĻŋāϤ⧇ āϚāϞāĻŽāĻžāύ āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁āϰ āĻŽā§‹āϟ āĻļāĻ•ā§āϤāĻŋ E, āĻ•āĻŽā§āĻĒāĻžāĻ™ā§āĻ• āĻ…āĻĒāϰāĻŋāĻŦāĻ°ā§āϤāĻŋāϤ āϰ⧇āϖ⧇ āĻŦāĻŋāĻ¸ā§āϤāĻžāϰ āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻ•āϰāϞ⧇ āϏāϰāϞ āĻ›āĻ¨ā§āĻĻāĻŋāϤ āĻ—āϤāĻŋāϤ⧇ āϚāϞāĻŽāĻžāύ āĻŦāĻ¸ā§āϤ⧁āϟāĻŋāϰ āĻŽā§‹āϟ āĻ—āϤāĻŋāĻļāĻ•ā§āϤāĻŋ āĻ•āϤ āĻšāĻŦ⧇?

 

āĻ•. E

āĻ–. 2E

āĻ—. E/2

āϘ. 4E

 

āωāĻ¤ā§āϤāϰ :  āϘ. 4E

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

āĻ—āϤāĻŋ āĻļāĻ•ā§āϤāĻŋ, k = ÂŊ K (A2 – x2), āĻ…āĻ°ā§āĻĨāĻžā§Ž āĻ—āϤāĻŋāĻļāĻ•ā§āϤāĻŋ āĻŦāĻŋāĻ¸ā§āϤāĻžāϰ⧇āϰ āĻŦāĻ°ā§āϗ⧇āϰ āϏāĻŽāĻžāύ⧁āĻĒāĻžāϤāĻŋāĻ•

∴ A āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻ•āϰāĻž āĻšāϞ⧇ āĻŽā§‹āϟ āĻ—āϤāĻŋāĻļāĻ•ā§āϤāĻŋ (22)xE= 4E

 

⧧⧍. āĻšāĻžāχāĻĄā§āϰ⧋āĻœā§‡āύ āĻĒāϰāĻŽāĻžāϪ⧁āϰ āĻļāĻ•ā§āϤāĻŋ āĻ¸ā§āϤāϰ⧇āϰ āĻĒā§āϰāĻ•āĻžāĻļ $\mathrm{E}=-\frac{13.5}{n^{2}} \mathrm{eV},(\mathrm{n}=1.2, \ldots)$. āĻ­ā§‚āĻŽāĻŋ āĻ…āĻŦāĻ¸ā§āĻĨāĻž āĻĨ⧇āϕ⧇ āĻĒāϰāĻŦāĻ°ā§āϤ⧀ āωāĻšā§āϚāϤāϰ āĻļāĻ•ā§āϤāĻŋāĻ¸ā§āϤāϰ⧇ āϝ⧇āϤ⧇ āĻāĻ•āϟāĻŋ āĻšāĻžāχāĻĄā§āϰ⧋āĻœā§‡āύ āĻĒāϰāĻŽāĻžāϪ⧁ āĻ•āĻŋ āĻĒāϰāĻŋāĻŽāĻžāĻŖ āĻļāĻ•ā§āϤāĻŋ āĻļā§‹āώāĻŖ āĻ•āϰ⧇?

 

āĻ•. 3.4 eV

āĻ–. 4.5 eV

āĻ—. 10.2 eV

āϘ. 13.6 eV

 

āωāĻ¤ā§āϤāϰ : āĻ•. 3.4 eV

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

n = 2 āĻšāĻ˛ā§‡Â $\mathrm{E}=\frac{13.5}{2^{2}} \mathrm{eV}=3 \cdot 4 \mathrm{eV}$

 

ā§§ā§Š. āĻāĻ•āϟāĻŋ āĻ•āĻŖāĻžāϰ āĻ­āϰāĻŦ⧇āĻ— P, āĻ•āĻŖāĻžāϟāĻŋāϰ āĻ—āϤāĻŋāĻļāĻ•ā§āϤāĻŋ āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻ•āϰāĻž āĻšāϞ⧇ āĻāϰ āύāϤ⧁āύ āĻ­āϰāĻŦ⧇āĻ— āĻ•āϤ āĻšāĻŦ⧇?

 

āĻ•. 2P

āĻ–. 2P

āĻ—. 4P

āϘ. 8P

 

āωāĻ¤ā§āϤāϰ : āĻ•. 2P

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{E}=\frac{1}{2} \mathrm{x} \frac{p^{2}}{m}, \mathrm{E} \propto \mathrm{P}^{2} \therefore 2 \mathrm{E}$ āĻāϰ āϜāĻ¨ā§āϝ āύāϤ⧁āύ āĻ­āϰāĻŦ⧇āĻ— (√2P)

 

ā§§ā§Ē. 100 W āĻ•ā§āώāĻŽāϤāĻž āϏāĻŽā§āĻĒāĻ¨ā§āύ āĻāĻ•āϟāĻŋ āĻšāĻŋāϟāĻžāϰ⧇ 2 kg āĻ­āϰ⧇āϰ āĻāĻ•āϟāĻŋ āĻ•āĻĒāĻžāϰ⧇āϰ āĻ–āĻŖā§āĻĄāϕ⧇ 40 s āϤāĻžāĻĒ āĻĻā§‡ā§ŸāĻž āĻšāϞ⧇ āĻ–āĻŖā§āĻĄāϟāĻŋāϰ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻž āĻ•āϤ āĻŦ⧃āĻĻā§āϧāĻŋ āĻšāĻŦ⧇? āĻ•āĻĒāĻžāϰ⧇āϰ āφāĻĒ⧇āĻ•ā§āώāĻŋāĻ• āϤāĻžāĻĒ 400 J/(Kg K)

 

āĻ•. 5 K

āĻ–. 10 K

āĻ—. 20 K

āϘ. 50 K

 

āωāĻ¤ā§āϤāϰ : āĻ•. 5 K

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$\mathrm{Q}=\mathrm{Pt}=100 \times 40=4000, \therefore \Delta \theta=\frac{4000}{2 \times 400}=5 \mathrm{~K}$

 

ā§§ā§Ģ. 300 Hz āĻ•āĻŽā§āĻĒāĻžāĻ™ā§āϕ⧇āϰ āĻāĻŦāĻ‚ āĻŦāĻŋāĻĒāϰ⧀āϤ āĻĻāĻŋāϕ⧇ āĻ…āĻ—ā§āϰāĻ—āĻžāĻŽā§€ āĻĻ⧁āϟāĻŋ āĻ…āĻ­āĻŋāĻ¨ā§āύ āϤāϰāĻ™ā§āϗ⧇āϰ āωāĻĒāϰāĻŋāĻĒāĻžāϤāύ⧇āϰ āĻĢāϞ⧇ āĻāĻ•āϟāĻŋ āĻ¸ā§āĻĨāĻŋāϰ āϤāϰāĻ™ā§āϗ⧇āϰ āϏ⧃āĻˇā§āϟāĻŋ āĻšā§Ÿā§‡āϛ⧇āĨ¤ āĻ¸ā§āĻĨāĻŋāϰ āϤāϰāĻ™ā§āϗ⧇āϰ āĻĒāϰ āĻĒāϰ āĻĻ⧁āϟāĻŋ āύāĻŋāĻ¸ā§āĻĒāĻ¨ā§āĻĻ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϰ āĻĻā§‚āϰāĻ¤ā§āĻŦ 1.5 m. āĻ…āĻ—ā§āϰāĻ—āĻžāĻŽā§€ āϤāϰāĻ™ā§āĻ— āĻĻ⧁āϟāĻŋāϰ āĻŦ⧇āĻ— āĻ•āϤ?

 

āĻ•. 100 m/s

āĻ–. 200 m/s

āĻ—. 450 m/s

āϘ. 900 m/s

 

āωāĻ¤ā§āϤāϰ : āϘ. 900 m/s

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

          

 

ā§§ā§Ŧ. āĻāĻ•āϟāĻŋ āĻ•ā§ˆāĻļāĻŋāĻ• āύāϞ⧇āϰ āĻŦā§āϝāĻžāϏ 0.04x10-4m. āĻāϰ āĻāĻ• āĻĒā§āϰāĻžāĻ¨ā§āϤ āĻĒāĻžāύāĻŋāϤ⧇ āĻĄā§āĻŦāĻžāϞ⧇ āĻĒāĻžāύāĻŋ āύāϞ⧇āϰ āĻ­āĻŋāϤāϰ 0.082 m āωāĻĒāϰ⧇ āωāϠ⧇āĨ¤ āĻĒāĻžāύāĻŋāϰ āϤāϞ āϟāĻžāύ āĻ•āϤ? āĻĻ⧇āĻ“ā§ŸāĻž āφāϛ⧇, āĻ¸ā§āĻĒāĻ°ā§āĻļ āϕ⧋āĻŖ = 0°C āĻāĻŦāĻ‚ āĻĒāĻžāύāĻŋāϰ āϘāύāĻ¤ā§āĻŦ = 1.0x103 Kg/m3

 

āĻ•. 8.5 x 10-4 N/m

āĻ–. 7.5 x 10-4 N/m

āĻ—. 9.0 x 10-4 N/m

āϘ. 8.0 x 10-4 N/m

 

āωāĻ¤ā§āϤāϰ : āϘ. 8.0 x 10-4 N/m

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{T}=\frac{\mathrm{hr} \rho \mathrm{g}}{2 \cos \theta}, \mathrm{T}=\frac{0.082 \times 0.02 \times 10^{-4} \times 10^{3} \times 9.8}{2}=8.0 \times 10^{-4}$

 

 

ā§§ā§­. āĻāĻ•āϟāĻŋ āϧāĻžāϰāϕ⧇āϰ āĻĻ⧁āχ āĻĒāĻžāϤ⧇āϰ āĻŽāĻ§ā§āϝ⧇ āĻŦāĻŋāĻ­āĻŦ āĻĒāĻžāĻ°ā§āĻĨāĻ•ā§āϝ V āĻāĻŦāĻ‚ āϧāĻžāϰāϕ⧇āϰ āϏāĻžā§āϚāĻŋāϤ āĻļāĻ•ā§āϤāĻŋ X, āϧāĻžāϰāϕ⧇āϰ āĻŦāĻŋāĻ­āĻŦ āĻĒāĻžāĻ°ā§āĻĨāĻ•ā§āϝ āĻŦ⧃āĻĻā§āϧāĻŋ āĻ•āϰ⧇ 3V āĻ•āϰāĻž āĻšāϞ⧇ āϏāĻžā§āϚāĻŋāϤ āĻļāĻ•ā§āϤāĻŋ āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒā§‡ā§Ÿā§‡ āĻ•āϤ āĻšāĻŦ⧇?

 

āĻ•. 3 X

āĻ–. 6 X

āĻ—. 9 X

āϘ. 27 X

 

āωāĻ¤ā§āϤāϰ : āĻ—. 9 X

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}, 3 \mathrm{~V}$ āĻ•āϰāĻž āĻšāϞ⧇ āϏāĻžā§āϚāĻŋāϤ āĻļāĻ•ā§āϤāĻŋ āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒā§‡ā§Ÿā§‡ āĻšāĻŦ⧇ $9X$

 

 

ā§§ā§Ž. āĻāĻ•āϟāĻŋ ${ }^{238}{ }_{93} \mathrm{~U}$  āύāĻŋāωāĻ•ā§āϞāĻŋ⧟āĻžāϏ āĻĻ⧁āχ āϧāĻžāĻĒ⧇ āĻ•ā§āώ⧟ āĻšā§Ÿā§‡ ${ }^{234}{ }_{91} \mathrm{~Pa}$ āύāĻŋāωāĻ•ā§āϞāĻŋ⧟āĻžāϏ āϏ⧃āĻˇā§āϟāĻŋ āĻ•āϰ⧇āĨ¤ āĻāχ āĻĻ⧁āχ āϧāĻžāĻĒ⧇ āϕ⧀ āϕ⧀ āϧāϰāύ⧇āϰ āϰāĻļā§āĻŽāĻŋ āύāĻŋāĻ°ā§āĻ—āϤ āĻšā§Ÿ?

 

āĻ•. Îą and ß

āĻ–. Îą and Îŗ

āĻ—. ß and ß

āϘ. ß and Îŗ

 

āωāĻ¤ā§āϤāϰ :

 

⧧⧝. āĻāĻ•āϟāĻŋ āĻ•āĻŽāύ āĻāĻŽāĻŋāϟāĻžāϰ āĻŸā§āϰāĻžāύāϜāĻŋāĻ¸ā§āϟāĻžāϰ⧇āϰ ß = 100 āĻāĻŦāĻ‚ IB = 50 ÂĩA āĻšāϞ⧇ Îą āĻ•āϤ?

 

āĻ•. 1.01

āĻ–. 0.99

āĻ—. 1.00

āϘ. 1.10

 

āωāĻ¤ā§āϤāϰ :  āĻ–. 0.99

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$\beta=\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}} \therefore \mathrm{I}_{\mathrm{C}}=100 \times 50 \times 10^{-6}=0.005, \therefore \mathrm{I}_{\mathrm{E}}=0.00005+0.005=0.00505$,

$\therefore \alpha=\frac{0.005}{0.00505}=0.99$

 

⧍ā§Ļ. āĻāĻ•āϟāĻŋ āĻ—āĻžā§œāĻŋ āϏ⧋āϜāĻž āωāĻ¤ā§āϤāϰ āĻĻāĻŋāϕ⧇ 90 m āĻĒāĻĨ 15 s āϏāĻŽā§Ÿā§‡ āĻ…āϤāĻŋāĻ•ā§āϰāĻŽ āĻ•āϰ⧇āĨ¤ āĻĒāϰāĻŦāĻ°ā§āϤ⧀āϤ⧇ āĻ—āĻžā§œāĻŋāϟāĻŋ āĻĻā§āϰ⧁āϤ āϘ⧁āϰ⧇ āĻĻāĻ•ā§āώāĻŋāĻŖ āĻĻāĻŋāϕ⧇ 40 m āĻĻā§‚āϰāĻ¤ā§āĻŦ 5 s āϏāĻŽā§Ÿā§‡ āĻ…āϤāĻŋāĻ•ā§āϰāĻŽ āĻ•āϰ⧇āĨ¤ āĻāχ 20 s āϏāĻŽā§ŸāĻ•āĻžāϞ⧇ āĻ—āĻžā§œāĻŋāϟāĻŋāϰ āĻ—ā§œ āĻŦ⧇āϗ⧇āϰ āĻŽāĻžāύ āĻ•āϤ?

 

āĻ•. 2.5 m/s

āĻ–. 5.0 m/s

āĻ—. 6.5 m/s

āϘ. 7.0 m/s

 

āωāĻ¤ā§āϤāϰ :  āĻ—. 6.5 m/s

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

130/20 = 6.5

 

⧍⧧. āĻāĻ•āϟāĻŋ āϤāĻžāϰ⧇āϰ āωāĻĒāϰ āϟāĻžāύ F āĻšāϞ⧇ āĻĻ⧈āĻ°ā§āĻ˜ā§āϝāĻŦ⧃āĻĻā§āĻ­āĻŋ āĻšā§Ÿ x, āϤāĻžāϰāϟāĻŋ āϝāĻĻāĻŋ āĻšā§āϕ⧇āϰ āϏ⧂āĻ¤ā§āϰ āĻŽā§‡āύ⧇ āϚāϞ⧇ āĻāĻŦāĻ‚ āϤāĻžāϰ⧇āϰ āωāĻĒāĻžāĻĻāĻžāύ⧇āϰ āĻ‡ā§ŸāĻ‚ āϗ⧁āĻŖāĻžāĻ™ā§āĻ• Y āĻšā§Ÿ āϤāĻŦ⧇ āϤāĻžāϰ⧇ āϏāĻžā§āϚāĻŋāϤ āĻŦāĻŋāĻ­āĻŦ āĻļāĻ•ā§āϤāĻŋ āĻ•āϤ?

 

A. $\frac{1}{2} \mathrm{Yx}$
B. $\mathrm{Yx}$
C. $\frac{1}{2} \mathrm{Fx}$
D. $\mathrm{Fx}$

 

āωāĻ¤ā§āϤāϰ :  C. $\frac{1}{2} \mathrm{Fx}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

                 $w=\frac{1}{2} \times \frac{Y A l^{2}}{L}$ , āĻāĻ–āĻžāύ⧇, $\mathrm{Y}=\frac{F L}{A x}$  āĻŦāĻž $\frac{A}{\mathrm{~L}}=\frac{F}{Y x}, \mathrm{~W}=\frac{1}{2} \mathrm{x} \frac{Y F x^{2}}{Y x}=\frac{1}{2} \mathrm{Fx}$

 

⧍⧍.  āϤ⧋āĻŽāĻžāϰ āĻāĻ•āϟāĻŋ 15Ί āϰ⧋āϧ āĻĒā§āĻ°ā§Ÿā§‹āϜāύ āĻ•āĻŋāĻ¨ā§āϤ⧁ āϤ⧋āĻŽāĻžāϰ āĻ•āĻžāϛ⧇ āĻ•ā§Ÿā§‡āĻ•āϟāĻŋ 10Ί āϰ⧋āϧ āφāϛ⧇āĨ¤ āϕ⧀āĻ­āĻžāĻŦ⧇ āϤ⧁āĻŽāĻŋ 10Ί āϰ⧋āϧ āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻ•āϰ⧇ 15Ί āϰ⧋āϧāϟāĻŋ āϤ⧈āϰāĻŋ āĻ•āϰāĻŦ⧇?

 

āĻ•. āϤāĻŋāύāϟāĻŋ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āϏāĻ‚āϝ⧋āϗ⧇

āĻ–. āϤāĻŋāύāϟāĻŋ āĻļā§āϰ⧇āĻŖā§€āĻŦāĻĻā§āϧ āϏāĻ‚āϝ⧋āϗ⧇

āĻ—. āĻĻ⧁āϟāĻŋ āĻļā§āϰ⧇āĻŖā§€āĻŦāĻĻā§āϧ āϏāĻ‚āϝ⧋āϗ⧇

āϘ. āĻĻ⧁āϟāĻŋ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āϏāĻ‚āϝ⧋āϗ⧇ āĻāĻŦāĻ‚ āĻāĻ•āϟāĻŋ āĻļā§āϰ⧇āĻŖāĻŋāĻŦāĻĻā§āϧ āϏāĻ‚āϝ⧋āϗ⧇

 

āωāĻ¤ā§āϤāϰ :  āϘ. āĻĻ⧁āϟāĻŋ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āϏāĻ‚āϝ⧋āϗ⧇ āĻāĻŦāĻ‚ āĻāĻ•āϟāĻŋ āĻļā§āϰ⧇āĻŖāĻŋāĻŦāĻĻā§āϧ āϏāĻ‚āϝ⧋āϗ⧇

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$10+\frac{1}{10^{-1}+10^{-1}}=15$

 

ā§¨ā§Š. āĻŦāϞāĻŦāĻŋāĻĻā§āϝāĻžāϰ āĻŦāĻŋāĻ­āĻŋāĻ¨ā§āύ āĻŽā§ŒāϞāĻŋāĻ• āϭ⧌āϤ āϰāĻžāĻļāĻŋ āϏāĻŽā§‚āĻš āĻšāϞ?

 

āĻ•. āĻ­āϰ, āĻŦāϞ āĻāĻŦāĻ‚ āϏāĻŽā§Ÿ

āĻ–. āĻ­āϰ, āĻĻ⧈āĻ°ā§āĻ˜ā§āϝ āĻāĻŦāĻ‚ āϏāĻŽā§Ÿ

āĻ—. āĻŦāϞ, āĻļāĻ•ā§āϤāĻŋ āĻāĻŦāĻ‚ āϏāĻŽā§Ÿ

āϘ. āĻŦāϞ, āĻ­āϰ āĻāĻŦāĻ‚ āϏāĻŽā§Ÿ

 

āωāĻ¤ā§āϤāϰ :  āĻ–. āĻ­āϰ, āĻĻ⧈āĻ°ā§āĻ˜ā§āϝ āĻāĻŦāĻ‚ āϏāĻŽā§Ÿ

 

⧍ā§Ē. āĻāĻ•āχ āĻĻ⧈āĻ°ā§āĻ˜ā§āϝ āĻāĻŦāĻ‚ āĻāĻ•āχ āĻĒāĻĻāĻžāĻ°ā§āĻĨ āĻĻāĻŋā§Ÿā§‡ āϤ⧈āϰāĻŋ āĻĻ⧁āϟāĻŋ āϤāĻžāϰ P āĻāĻŦāĻ‚ Q āϕ⧇ āĻāĻ•āϟāĻŋ āĻŦā§āϝāĻžāϟāĻžāϰāĻŋāϰ āϏāĻžāĻĨ⧇ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ⧇ āϏāĻ‚āϝ⧁āĻ•ā§āϤ āĻ•āϰāĻž āĻšā§Ÿā§‡āϛ⧇āĨ¤ P āϤāĻžāϰ⧇āϰ āĻŦā§āϝāĻžāϏ 2mm āĻāĻŦāĻ‚ Q āϤāĻžāϰ⧇āϰ āĻŦā§āϝāĻžāϏ 1mm. P āĻāĻŦāĻ‚ Q āĻāϰ āϤ⧜āĻŋā§Ž āĻĒā§āϰāĻŦāĻžāĻšā§‡āϰ āĻ…āύ⧁āĻĒāĻžāϤ āĻ•āϤ?

 

āĻ•. 1/4

āĻ–. 1/2

āĻ—. 2/1

āϘ. 4/1

 

āωāĻ¤ā§āϤāϰ :  āϘ. 4/1

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\frac{l_{1}}{l_{2}}=\frac{R_{2}}{R_{1}}=\frac{A_{1}}{A_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{4}{1}$

 

⧍ā§Ģ. 5 kg āĻ­āϰ⧇āϰ āĻāĻ•āϟāĻŋ āϰāĻžāχāĻĢ⧇āϞ āĻĨ⧇āϕ⧇ 20 g āĻ­āϰ⧇āϰ āĻāĻ•āϟāĻŋ āĻŦ⧁āϞ⧇āϟ 1000 m/s āĻ—āϤāĻŋāϤ⧇ āϛ⧁āĻŸā§‡ āϝāĻžā§ŸāĨ¤ āĻĒāĻŋāĻ›āύ āĻĻāĻŋāϕ⧇ āϰāĻžāχāĻĢ⧇āϞ⧇āϰ āϧāĻžāĻ•ā§āĻ•āĻžāϰ āĻŦ⧇āĻ— āĻ•āϤ?

 

āĻ•. 4 m/s

āĻ–. 4000 m/s

āĻ—. 400 m/s

āϘ. 40 m/s

 

āωāĻ¤ā§āϤāϰ :  āĻ•. 4 m/s

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{V}=\frac{0.02 \times 1000}{5}=4$

 

⧍ā§Ŧ. āĻāĻ•āϟāĻŋ āĻĒāĻžāĻĨāϰāϕ⧇ āĻ¸ā§āĻĨāĻŋāϰ āĻ…āĻŦāĻ¸ā§āĻĨāĻžā§Ÿ āĻāĻ•āϟāĻŋ āωāρāϚ⧁ āĻĻāĻžāϞāĻžāύ āĻĨ⧇āϕ⧇ āĻ›ā§‡ā§œā§‡ āĻĻ⧇āĻ“ā§ŸāĻž āĻšāϞāĨ¤ āĻ­ā§‚āĻŽāĻŋāϤ⧇ āĻĒ⧌āĻ›āĻžāϤ⧇ āĻĒāĻžāĻĨāϰāϟāĻŋāϰ 4 s āĻāϰ āĻŦ⧇āĻļāĻŋ āϏāĻŽā§Ÿ āϞāĻžāϗ⧇āĨ¤ āĻŦāĻžāϤāĻžāϏ⧇āϰ āϘāĻ°ā§āώāĻŖ āĻ•ā§āώ⧁āĻĻā§āϰ āĻšāϞ⧇ āĻĒāĻžāĻĨāϰāϟāĻŋāϰ āĻĒā§āϰāĻĨāĻŽ 4 s āϏāĻŽā§Ÿā§‡ āĻĒāϤāύ⧇āϰ āĻĻā§‚āϰāĻ¤ā§āĻŦ āĻāĻŦāĻ‚ āĻĒā§āϰāĻĨāĻŽ 2 s āϏāĻŽā§Ÿā§‡ āĻĒāϤāύ⧇āϰ āĻĻā§‚āϰāĻ¤ā§āĻŦ⧇āϰ āĻ…āύ⧁āĻĒāĻžāϤ āĻ•āϤ?

 

āĻ•. 1/4

āĻ–. 4/1

āĻ—. 1/2

āϘ. 2/1

 

āωāĻ¤ā§āϤāϰ :  āĻ–. 4/1

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2}, \therefore \frac{h_{1}}{h_{2}}=\left(\frac{t_{1}}{t_{2}}\right)^{2}=\frac{4}{1}$

 

⧍⧭. āϚāĻŋāĻ¤ā§āϰ⧇ āĻāĻ•āϟāĻŋ āĻŦāĻ°ā§āϤāύ⧀āϤ⧇ āϏāĻŽāĻžāĻ¨ā§āϤāϰāĻžāϞ āϏāĻ¨ā§āύāĻŋāĻŦ⧇āĻļ⧇ āϏāĻ‚āϝ⧁āĻ•ā§āϤ āϤāĻŋāύāϟāĻŋ āϰ⧋āϧ āĻĻ⧇āĻ–āĻžāύ⧋ āĻšā§Ÿā§‡āϛ⧇āĨ¤ āĻŦā§āϝāĻžāϟāĻžāϰāĻŋāϰ āϤ⧜āĻŋā§Ž-āϚāĻžāϞāĻ• āĻļāĻ•ā§āϤāĻŋ  12 V āĻāĻŦāĻ‚ āĻ…āĻ­ā§āϝāĻ¨ā§āϤāϰ⧀āĻŖ āϰ⧋āϧ āύāĻ—āĻ¨ā§āϝāĨ¤ āĻ…ā§āϝāĻžāĻŽāĻŋāϟāĻžāϰ⧇āϰ āĻĒāĻžāĻ  3.2 A āĻšāϞ⧇, X āĻāϰ āϰ⧋āϧ āĻ•āϤ?

 

āĻĸāĻžāĻ•āĻž āĻŦāĻŋāĻļā§āĻŦāĻŦāĻŋāĻĻā§āϝāĻžāϞāϝāĻŧ 'āĻ•' āχāωāύāĻŋāϟ āĻĒā§āϰāĻļā§āύ (⧍ā§Ļ⧧⧍ - ⧍ā§Ļā§§ā§Š)-1

 

āĻ•. 2.1 Ί

āĻ–. 4.6 Ί

āĻ—. 6.0 Ί

āϘ. 15 Ί

 

āωāĻ¤ā§āϤāϰ :  āϘ. 15 Ί

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{R}=\frac{12}{3.2}=3.75$ , āφāĻŦāĻžāĻ°Â $\frac{1}{10^{-1}+10^{-1}+15^{-1}}=3.75$

 

ā§¨ā§Ž. āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁āϕ⧇ āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāϰ āϏāĻžāĻĨ⧇ 30° āϕ⧋āϪ⧇ āύāĻŋāĻ•ā§āώ⧇āĻĒ āĻ•āϰāĻž āĻšāϞāĨ¤ āĻĒāϰāĻŦāĻ°ā§āϤ⧀āϤ⧇ āĻāĻ•āχ āĻŦāĻ¸ā§āϤ⧁āϕ⧇ āĻāĻ•āχ āφāĻĻāĻŋ āĻĻā§āϰ⧁āϤāĻŋāϤ⧇ āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāϰ āϏāĻžāĻĨ⧇ 40° āϕ⧋āϪ⧇ āύāĻŋāĻ•ā§āώ⧇āĻĒ āĻ•āϰāĻž āĻšāϞāĨ¤ āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύāϟāĻŋ āϏāĻ¤ā§āϝ āύ⧟

 

āĻ•. āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāĻ• āĻĒāĻžāĻ˛ā§āϞāĻž āĻŦ⧃āĻĻā§āĻ­āĻŋ āĻĒ⧇āϞ

āĻ–. āĻŦ⧇āϗ⧇āϰ āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāĻ• āωāĻĒāĻžāĻ‚āĻļ āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒ⧇āϞ

āĻ—. āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āωāĻšā§āϚāϤāĻž āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒ⧇āϞ

āϘ. āĻŦāĻ¸ā§āϤ⧁āϟāĻŋāϰ āωāĻĄā§āĻĄā§ŸāύāĻ•āĻžāϞ āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒ⧇āϞ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āĻŦ⧇āϗ⧇āϰ āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāĻ• āωāĻĒāĻžāĻ‚āĻļ āĻŦ⧃āĻĻā§āϧāĻŋ āĻĒ⧇āϞ

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

āĻŦ⧇āϗ⧇āϰ āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāĻ• āωāĻĒāĻžāĻ‚āĻļ, $\mathrm{V}_{0} \cos \theta_{0}$ , āĻāĻŦāĻ‚Â $\cos 30^{\circ}>\cos 40^{\circ}$

 

⧍⧝. āĻāĻ•āϟāĻŋ āφāĻĻāĻ°ā§āĻļ āĻŸā§āϰāĻžāĻ¨ā§āϏāĻĢāĻ°ā§āĻŽāĻžāϰ⧇āϰ āĻ—ā§ŒāĻŖ āĻ“ āĻŽā§‚āĻ–ā§āϝ āϕ⧁āĻŖā§āĻĄāϞ⧀āϰ āĻĒāĻžāĻ• āϏāĻ‚āĻ–ā§āϝāĻžāϰ āĻ…āύ⧁āĻĒāĻžāϤ 6:1. āϝāĻĻāĻŋ āĻŽā§āĻ–ā§āϝ āϕ⧁āĻŖā§āĻĄāϞ⧀āϤ⧇ āĻĒā§āϰāϤāĻŋ āϏ⧇āϕ⧇āĻ¨ā§āĻĄā§‡ āĻŦā§āϝ⧟āĻŋāϤ āĻļāĻ•ā§āϤāĻŋ 6 J āĻšā§Ÿ, āϤāĻŦ⧇ āĻ—ā§ŒāĻŖ āϕ⧁āĻŖā§āĻĄāϞ⧀āϤ⧇ āĻŦ⧈āĻĻā§āϝ⧁āϤāĻŋāĻ• āĻ•ā§āώāĻŽāϤāĻž āĻ•āϤ?

 

āĻ•. 6 J

āĻ–. 36 Js-1

āĻ—. 6 W

āϘ. 36 W

 

āωāĻ¤ā§āϤāϰ :  āĻ—. 6 W

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

āĻŽā§āĻ–ā§āϝ āĻ“ āĻ—ā§ŒāĻŖ āϕ⧁āĻŖā§āĻĄāϞ⧀āϤ⧇ āĻŦ⧈āĻĻā§āϝ⧁āϤāĻŋāĻ• āĻ•ā§āώāĻŽāϤāĻž āϏāĻŽāĻžāύāĨ¤

 

ā§Šā§Ļ. m āĻ­āϰ⧇āϰ āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁ r āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ⧇āϰ āĻŦ⧃āĻ¤ā§āϤāĻžāĻ•āĻžāϰ āĻĒāĻĨ⧇ āϏāĻŽāĻĻā§āϰ⧁āϤāĻŋāϤ⧇ āϚāϞāϛ⧇āĨ¤ āĻŦ⧃āϤāĻžāĻ•āĻžāϰ āĻ—āϤāĻŋāϰ āĻĒāĻ°ā§āϝāĻžā§ŸāĻ•āĻžāϞ T, āĻŦāĻ¸ā§āϤ⧁āϟāĻŋāϰ āωāĻĒāϰ āϕ⧇āĻ¨ā§āĻĻā§āϰāĻŽā§āĻ–ā§€ āĻŦāϞ⧇āϰ āĻŽāĻžāύ āĻ•āϤ?

āĻ•. 4Ī€2mr/T2

āĻ–. 4Ī€2mr/T

āĻ—. 4Ī€mr2/T2

āϘ. Ī€mr2

āωāĻ¤ā§āϤāϰ : āĻ•. 4Ī€2mr/T2

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{F}=\mathrm{m} \alpha, \alpha=\omega^{2} \mathrm{r}=\left(\frac{2 \pi}{\tau}\right)^{2} \times r \therefore F=\frac{4 \pi^{2} m r}{\tau^{2}}$

 

āĻ—āĻŖāĻŋāϤ

 

1. y = mx, y = m1x āĻāĻŦāĻ‚ y = b āϏāϰāϞāϰ⧇āĻ–āĻžāĻ¤ā§āϰ⧟ āĻĻā§āĻŦāĻžāϰāĻž āĻ—āĻ āĻŋāϤ āĻ¤ā§āϰāĻŋāϭ⧁āĻœā§‡āϰ āĻŦāĻ°ā§āĻ—āĻāĻ•āϕ⧇ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻšāĻŦ⧇

 

(A) $\frac{\mathrm{b}^{2}\left(\mathrm{~m}_{1}-\mathrm{m}\right)}{2 \mathrm{~mm}_{1}}$

(B) $\frac{\mathrm{b}^{2}\left(\mathrm{~m}-\mathrm{m}_{1}\right)}{2 \mathrm{~mm}_{1}}$

(C) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{\mathrm{mm}_{1}}$

(D) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$

Answer: (D) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ : 

$\mathrm{O} \equiv(\mathrm{o}, \mathrm{o})$

$\mathrm{A} \equiv\left(\frac{\mathrm{b}}{\mathrm{m}}, \mathrm{b}\right)$

$\mathrm{B} \equiv\left(\frac{\mathrm{b}}{\mathrm{m}_{1}}, \mathrm{~b}\right)$

$\Delta \mathrm{OAB}=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ \frac{\mathrm{b}}{\mathrm{m}} & \mathrm{b} & 1 \\ \frac{\mathrm{b}}{\mathrm{m}_{1}} & \mathrm{~b} & 1\end{array}\right|=\frac{1}{2}\left(\frac{\mathrm{b}^{2}}{\mathrm{~m}}-\frac{\mathrm{b}^{2}}{\mathrm{~m}_{1}}\right)=\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$

Answer: (D) $\frac{\mathrm{b}^{2}\left|\mathrm{~m}-\mathrm{m}_{1}\right|}{2 \mathrm{~mm}_{1}}$

 

2. 3x2 + 5y2 = 15 āωāĻĒāĻŦ⧃āĻ¤ā§āϤ⧇āϰ āĻ‰ā§Žāϕ⧇āĻ¨ā§āĻĻā§āϰāĻŋāĻ•āϤāĻž āĻšāĻŦ⧇

 

(A) $\sqrt{3 / 5}$
(B) $\sqrt{5 / 3}$
(C) $\sqrt{2 / 5}$
(D) $\sqrt{5 / 2}$

Answer: (C) $\sqrt{2 / 5}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ : 

$3 x^{2}+5 y^{2}=15$

$\Rightarrow \frac{3 x^{2}}{15}+\frac{5 y^{2}}{15}=0$

$\Rightarrow \frac{x^{2}}{5}+\frac{y^{2}}{3}=0$

$\therefore a^{2}=5, b^{2}=3$

$\because a>b$

$\mathrm{e}^{2}=\frac{\mathrm{a}^{2}-\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{5-3}{5}=\frac{2}{5}$

$\therefore e=\sqrt{\frac{2}{5}}$

Answer: (C) $\sqrt{2 / 5}$

3.  $\left(2 x^{2}-\frac{1}{2 x^{3}}\right)^{10}$ āĻāϰ āĻŦāĻŋāĻ¸ā§āϤ⧃āϤāĻŋāϤ⧇ x-āĻŦāĻ°ā§āϜāĻŋāϤ āĻĒāĻĻāϟāĻŋ āĻ•āϤāϤāĻŽ āĻāĻŦāĻ‚ āĻāϰ āĻŽāĻžāύ āĻ•āϤ?

 

(A)  āĻĒāĻžā§āϚāĻŽ āĻāĻŦāĻ‚ 840

(B)  āϚāϤ⧁āĻ°ā§āĻĨ āĻāĻŦāĻ‚ 1920

(C)  āώāĻˇā§āϟ āĻāĻŦāĻ‚ 252

(D)  āϏāĻĒā§āϤāĻŽ āĻāĻŦāĻ‚ 30

 

Answer: (A)  āĻĒāĻžā§āϚāĻŽ āĻāĻŦāĻ‚ 840

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

(r + 1) āϤāĻŽ āĻĒāĻĻ āĻšāĻŦ⧇

${ }^{10} \mathrm{C}_{\mathrm{r}}\left(2 \mathrm{x}^{2}\right)^{10-\mathrm{r}}\left(-\frac{1}{2 \mathrm{x}^{3}}\right)^{\mathrm{r}}$

$=10 \mathrm{C}_{\mathrm{r}} \cdot 2^{10-\mathrm{r}} \cdot \mathrm{x}^{20-2 \mathrm{r}} \cdot(-1)^{\mathrm{r}} \cdot 2^{-\mathrm{r}} \cdot \mathrm{x}^{-3 \mathrm{r}}$

$=(-1)^{\mathrm{r}} \cdot{ }^{10} \mathrm{C}_{\mathrm{r}} \cdot 2^{10-2 \mathrm{r}} \cdot \mathrm{x}^{20-5 \mathrm{r}}$

 

āϝāĻĻāĻŋ āĻĒāĻĻāϟāĻŋ x āĻŦāĻ°ā§āϜāĻŋāϤ āĻšā§Ÿ āϤāĻŦ⧇

$20-5 r=0 \Rightarrow 5 r=20 \Rightarrow r=4$

∴ (r + 1) = 5-āϤāĻŽ āĻĒāĻĻ x āĻŦāĻ°ā§āϜāĻŋāϤ āĻāĻŦāĻ‚ āĻāϰ āĻŽāĻžāύ āĻšāĻŦ⧇

$(-1)^{4} \cdot{ }^{10} \mathrm{C}_{4} \cdot 2^{2}=840$

Answer: (A) āĻĒāĻžā§āϚāĻŽ āĻāĻŦāĻ‚ 840

 

4. āĻāĻ•āϟāĻŋ āĻŦāĻ¸ā§āϤ⧁āϰ āĻ–āĻžā§œāĻž āωāĻĒāϰ⧇āϰ āĻĻāĻŋāϕ⧇ āĻĒā§āϰāĻ•ā§āώ⧇āĻĒāύ āĻ•āϰāϞ⧇ āύāĻŋāĻ°ā§āĻĻāĻŋāĻˇā§āϟ āĻŦāĻŋāĻ¨ā§āĻĻ⧁ P-āϤ⧇ āĻĒ⧌āĻ›āĻžāϤ⧇ t1 āϏāĻŽā§Ÿ āϞāĻžāϗ⧇āĨ¤ āϝāĻĻāĻŋ āφāϰāĻ“ t2 āϏāĻŽā§Ÿ āĻĒāϰ āĻŦāĻ¸ā§āϤ⧁āϟāĻŋ āĻ­ā§‚āĻŽāĻŋāϤ⧇ āĻĒāϤāĻŋāϤ āĻšā§Ÿ āϤāĻŦ⧇ āĻ•āĻŖāĻžāϟāĻŋāϰ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āωāĻšā§āϚāϤāĻž āĻšāĻŦ⧇

 

(A) $\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$

(B) $\frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$

(C) $\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}{ }^{2}\right)$

(D) $\frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}{ }^{2}+\mathrm{t}_{2}{ }^{2}\right)$

Answer: (B) $\frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

āĻŽā§‹āϟ āĻŦāĻŋāϚāϰāĻŖāĻ•āĻžāϞ = T = t1 + t2

āφāĻŦāĻžāϰ,

$\mathrm{T}=\frac{2 \mathrm{u}}{\mathrm{g}}$
$\Rightarrow \mathrm{u}=\frac{1}{2} \mathrm{~g} \mathrm{~T}=\frac{1}{2} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)$
$\Rightarrow \mathrm{u}^{2}=\frac{1}{4} \mathrm{~g}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$

∴ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āωāĻšā§āϚāϤāĻž āĻšāĻŦ⧇

$\frac{u^{2}}{2 g}=\frac{1}{8} g\left(t_{1}+t_{2}\right)^{2}$

Answer: $(\mathrm{B}) \frac{1}{8} \mathrm{~g}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}$

 

 

5.  $\tan \left(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{2}\right)\right)$ āĻāϰ āĻŽāĻžāύ āĻšāĻŦ⧇

 

(A) $\frac{5}{6}$
(B) $1$
(C) $\frac{\pi}{4}$
(D) $-\frac{5}{6}$

 

Answer: (B) $1$

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

āϏāĻžāϰāĻžāϏāϰāĻŋ Calculator āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻ•āϰ⧇:

āĻĸāĻžāĻ•āĻž āĻŦāĻŋāĻļā§āĻŦāĻŦāĻŋāĻĻā§āϝāĻžāϞāϝāĻŧ 'āĻ•' āχāωāύāĻŋāϟ āĻĒā§āϰāĻļā§āύ (⧍ā§Ļ⧧⧍ - ⧍ā§Ļā§§ā§Š)-2

 

6.  $\left|\begin{array}{lll}\mathrm{a} & 1 & \mathrm{~b}+\mathrm{c} \\ \mathrm{b} & 1 & \mathrm{c}+\mathrm{a} \\ \mathrm{c} & 1 & \mathrm{a}+\mathrm{b}\end{array}\right|$ āĻāϰ āĻŽāĻžāύ āĻšāĻŦ⧇

 

(A) 0
(B) $a b c(a+b)(b+c)(c+a)$
(C) abc
(D) $(\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c})(\mathrm{c}+\mathrm{a})$
Answer: (A) o

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\left|\begin{array}{lll}\mathrm{a} & 1 & \mathrm{~b}+\mathrm{c} \\ \mathrm{b} & 1 & \mathrm{c}+\mathrm{a} \\ \mathrm{c} & 1 & \mathrm{a}+\mathrm{b}\end{array}\right|$

$=\left|\begin{array}{lll}a+b+c & 1 & b+c \\ a+b+c & 1 & c+a \\ a+b+c & 1 & a+b\end{array}\right|$

$=a+b+c\left|\begin{array}{lll}1 & 1 & b+c \\ 1 & 1 & c+a \\ 1 & 1 & a+b\end{array}\right| \quad\left[c_{1}^{\prime}=c_{1}+c_{3}\right]$

$=0$

Answer: (A) $0$

 

7. 3P āĻāĻŦāĻ‚ 2P āĻŦāϞāĻĻā§āĻŦā§Ÿā§‡āϰ āϞāĻŦā§āϧāĻŋ RāĨ¤ āĻĒā§āϰāĻĨāĻŽ āĻŦāϞ āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻ•āϰāϞ⧇ āϞāĻŦā§āϧāĻŋāϰ āĻĒāϰāĻŋāĻŽāĻžāĻŖāĻ“ āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻšā§ŸāĨ¤ āĻŦāϞāĻĻā§āĻŦā§Ÿā§‡āϰ āĻ…āĻ¨ā§āϤāĻ°ā§āĻ—āϤ āϕ⧋āĻŖ āĻšāĻŦ⧇

 

(A) 110°

(B) 120°

(C) 150°

(D) 135°

 

Answer: (B) 120°

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

āĻŽāύ⧇ āĻ•āϰāĻŋ, 3P āĻāĻŦāĻ‚ 2P āĻŽāĻžāύ⧇āϰ āĻŦāϞāĻĻā§āĻŦā§Ÿā§‡āϰ āĻ…āĻ¨ā§āϤāĻ°ā§āĻ—āϤ āϕ⧋āĻŖ ÎąāĨ¤

$\therefore \mathrm{R}^{2}=9 \mathrm{P}^{2}+4 \mathrm{P}^{2}+12 \mathrm{P}^{2} \cos \alpha=13 \mathrm{P}^{2}+12 \mathrm{P}^{2} \cos \alpha$

āĻĒā§āϰāĻĨāĻŽ āĻŦāϞ⧇āϰ āĻŽāĻžāύ āĻĻā§āĻŦāĻŋāϗ⧁āĻŖ āĻ•āϰāϞ⧇

$4 \mathrm{R}^{2}=(6 \mathrm{P})^{2}+4 \mathrm{P}^{2}+2.2 \mathrm{P} .6 \mathrm{P} \cos \alpha=36 \mathrm{P}^{2}+4 \mathrm{P}^{2}+24 \mathrm{P}^{2} \cos \alpha=40 \mathrm{P}^{2}+24 \mathrm{P}^{2} \cos \alpha$

$\therefore \frac{4 \mathrm{R}^{2}}{\mathrm{R}^{2}}=\frac{\mathrm{P}^{2}(40+24 \cos \alpha)}{\mathrm{P}^{2}(13+12 \cos \alpha)}$

$\Rightarrow 52+48 \cos \alpha=40+24 \cos \alpha$

$\Rightarrow 24 \cos \alpha=-12$

$\Rightarrow \cos \alpha=-1 / 2$

$\therefore \alpha=120^{\circ}$

Answer: (B) $120^{\circ}$

 

8. āĻāĻ•āϟāĻŋ āχāϞ⧇āĻ•āĻŸā§āϰāĻŋāĻ• āĻĢāĻŋāĻ˛ā§āĻĄā§‡ āχāϞ⧇āĻ•āĻŸā§āϰāύ⧇āϰ āĻ¤ā§āĻŦāϰāĻŖ āĻāĻŦāĻ‚ āĻļāĻ•ā§āϤāĻŋ āϏāĻŽāĻžāύ⧁āĻĒāĻžāϤāĻŋāĻ•āĨ¤ $10^{-20} \mathrm{~N}$ āĻļāĻ•ā§āϤāĻŋāϰ āϜāĻ¨ā§āϝ āĻ¤ā§āĻŦāϰāĻŖ $10^{10} \frac{\mathrm{m}}{\mathrm{s}^{2}}$ āĻšāϞ⧇, $10^{-25} \mathrm{~N}$ āĻļāĻ•ā§āϤāĻŋāϰ āϜāĻ¨ā§āϝ āĻ¤ā§āĻŦāϰāĻŖ āĻšāĻŦ⧇

 

(A) $10^{5} \mathrm{~m} / \mathrm{s}^{2}$
(B) $10^{15} \mathrm{~m} / \mathrm{s}^{2}$
(C) $10^{-5-} \mathrm{m} / \mathrm{s}^{2}$
(D) $10^{-15} \mathrm{~m} / \mathrm{s}^{2}$

Answer: (A) $10^{5} \mathrm{~m} / \mathrm{s}^{2}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

āĻĻ⧇āĻ“ā§ŸāĻž āφāϛ⧇, āĻ¤ā§āĻŦāϰāĻŖ ∝ āĻļāĻ•ā§āϤāĻŋ

$10^{10} \mathrm{~m} / \mathrm{s}^{2} \propto 10^{-20} \mathrm{~N}$

āφāĻŦāĻžāϰ,

$x \propto 10^{-25}$

$\therefore \frac{x}{10^{10}}=\frac{10^{-25}}{10^{-20}}$

$\Rightarrow \mathrm{x}=10^{-25} \cdot 10^{10} \cdot 10^{20}=10^{5}$

Answer: (A) $10^{5} \mathrm{~m} / \mathrm{s}^{2}$

 

9. āĻĻāĻļāĻŽāĻŋāĻ• āϏāĻ‚āĻ–ā§āϝāĻž $2013$ āĻāϰ āĻĻā§āĻŦāĻŋāĻŽāĻŋāϕ⧇ āĻĒā§āϰāĻ•āĻžāĻļ āĻšāĻŦ⧇

 

(A) 11111011101

(B) 10111011111

(C) 10101110111

(D) 10101110101

 

Answer: (A) 11111011101

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$2 \mid 2012$
$2 \mid 1006-1$
$2 \mid 503-0$
$2 \mid 251-1$
$2 \mid 125-1$
$2 \mid 62-1$
$2 \mid 31-0$
$2 \mid 15-1$
$2 \mid 7-1$
$2 \mid 3-1$
$2 \mid 1-1$

  $0-1$
$\therefore(2013)_{10}=(11111011101)_{2}$

Answer: (A) 11111011101

 

10. x = y2 āĻāĻŦāĻ‚ y = x ‒ 2 āĻĻā§āĻŦāĻžāϰāĻž āφāĻŦāĻĻā§āϧ āĻ•ā§āώ⧇āĻ¤ā§āϰ⧇āϰ āĻ•ā§āώ⧇āĻ¤ā§āϰāĻĢāϞ āĻšāĻŦ⧇

(A) $1 \frac{1}{3}$

(B) $1 \frac{1}{3}$

(C) $4 \frac{1}{2}$

(D) $4 \frac{3}{4}$

Answer: (C) $4 \frac{1}{2}$

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{x}=\mathrm{y}^{2}$

$\mathrm{y}=\mathrm{x}-2 \Rightarrow \mathrm{x}=\mathrm{y}+2$

$\therefore \mathrm{y}^{2}=\mathrm{y}+2$

$\Rightarrow \mathrm{y}^{2}-\mathrm{y}-2=0$

$\Rightarrow \mathrm{y}^{2}-2 \mathrm{y}+\mathrm{y}-2=0$

$\Rightarrow \mathrm{y}(\mathrm{y}-2)+1(\mathrm{y}-2)=0$

$\Rightarrow(\mathrm{y}-2)(\mathrm{y}+1)=0$

$\Rightarrow \mathrm{y}=-1,2$

$\int_{-1}^{2}\left(\mathrm{y}+2-\mathrm{y}^{2}\right) \mathrm{dy}$

$=\int_{-1}^{2} \mathrm{y} \mathrm{dy}+2 \int_{-1}^{2} \mathrm{dy}-\int_{-1}^{2} \mathrm{y}^{2} \mathrm{dy}$

$=\left[\frac{1}{2} y^{2}\right]_{-1}^{2}+[2 y]_{-1}^{2}-\left[\frac{1}{3} y^{3}\right]_{-1}^{2}$

$=\frac{1}{2}(4-1)+2(2+1)-\frac{1}{3}(8+1)$

$=4 \frac{1}{2}$

Answer: (C) $4 \frac{1}{2}$

 

 

11.  āϝāĻĻāĻŋ $y=\sqrt{\cos 2 x}$ āĻšā§Ÿ, āϤāĻŦā§‡Â $\frac{d y}{d x}=$

 

(A) $-\frac{\sin 2 x}{\sqrt{\cos 2 x}}$

(B) $-\frac{\cos 2 x}{\sqrt{\sin 2 x}}$

(C) $-\frac{2 \sin 2 x}{\sqrt{\tan x}}$

(D) $\frac{\tan 2 \mathrm{x}}{\sqrt{\sin 2 \mathrm{x}}}$

Answer: (A) $-\frac{\sin 2 x}{\sqrt{\cos 2 x}}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$\mathrm{y}=\sqrt{\cos 2 \mathrm{x}}$ $\frac{d y}{d x}=\frac{1}{2 \sqrt{\cos 2 \mathrm{x}}} \frac{d}{d x} \cos 2 \mathrm{x}=\frac{1}{2 \sqrt{\cos 2 \mathrm{x}}}(-\sin 2 \mathrm{x}) \frac{d}{d x} 2 \mathrm{x}=-\frac{\sin 2 \mathrm{x}}{\sqrt{\cos 2 \mathrm{x}}}$ Answer: (A) $-\frac{\sin 2 \mathrm{x}}{\sqrt{\cos 2 \mathrm{x}}}$

 

12. āĻāĻ•āϜāύ āĻ•ā§ƒāώāĻ• āĻāĻ•āϟāĻŋ āĻ†ā§ŸāϤāĻžāĻ•āĻžāϰ āĻŦāĻžāĻ—āĻžāύ⧇āϰ āϤāĻŋāύ āĻĻāĻŋāĻ• āĻŦā§‡ā§œāĻž āĻĻāĻŋā§Ÿā§‡ āĻāĻŦāĻ‚ āϚāϤ⧁āĻ°ā§āĻĻāĻŋāϕ⧇ āĻāĻ•āϟāĻŋ āĻĻ⧇āĻ“ā§ŸāĻžāϞ āĻĻāĻŋā§Ÿā§‡ āĻ˜ā§‡āϰāĻžāĻ“ āĻĻāĻŋāϞāĨ¤ āϝāĻĻāĻŋ āϤāĻžāρāϰ āĻ•āĻžāϛ⧇ $100 m$ āĻŦā§‡ā§œāĻž āĻĨāĻžāϕ⧇ āϤāĻŦ⧇ āĻ˜ā§‡āϰāĻžāĻ“ āĻĻ⧇āĻ“ā§ŸāĻž āĻ¸ā§āĻĨāĻžāύ⧇āϰ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āĻ†ā§ŸāϤāύ āĻšāĻŦ⧇

 

(A)  2500 m2

(B)  1250 m2

(C)  750 m2

(D)  2000 m2

 

Answer: (B) 1250 m2

 

13. sin (ax + b) āĻāϰ n-āϤāĻŽ āĻ…āĻ¨ā§āϤāϰāĻ• āĻšāĻŦ⧇

 

(A) $a^{n} \sin \left(\frac{\pi}{2} n+a x+b\right)$

(B) $a^{n} \cos \left(\frac{\pi}{2} n+a x+b\right)$

(C) $(-1)^{\mathrm{n}} a^{\mathrm{n}} \sin (a x+b)$

(D) $(-1)^{\mathrm{n}} \mathrm{a}^{\mathrm{n}} \cos (\mathrm{ax}+\mathrm{b})$

Answer: (A) $a^{n} \sin \left(\frac{\pi}{2} n+a x+b\right)$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$y=\sin (a x+b)$

$y_{1}=a \cos (a x+b)=a \sin \left(\frac{\pi}{2}+a x+b\right)$

$y_{2}=a^{2} \cos \left(\frac{\pi}{2}+a x+b\right)=a^{2} \sin \left(2 \cdot \frac{\pi}{2}+a x+b\right)$

$y_{3}=a^{3} \cos \left(2 \cdot \frac{\pi}{2}+a x+b\right)=a^{3} \sin \left(3 \cdot \frac{\pi}{2}+a x+b\right)$

-------------------------------------------------

$\therefore \mathrm{y}_{\mathrm{n}}=\mathrm{a}^{\mathrm{n}} \sin \left(\frac{\pi}{2} \mathrm{n}+\mathrm{ax}+\mathrm{b}\right)$

Answer: $(A) a^{n} \sin \left(\frac{\pi}{2} n+a x+b\right)$

 

 

14.  ‘a’ āĻāϰ āϕ⧋āύ āĻŽāĻžāύ⧇āϰ āϜāĻ¨ā§āĻ¯Â  $2 \hat{\imath}+\hat{\jmath}-\hat{k}, 3 \hat{\imath}-2 \hat{\jmath}+4 \hat{k}$ āĻāĻŦāĻ‚Â $\hat{\imath}-3 \hat{\jmath}+a \hat{k}$ āϭ⧇āĻ•ā§āϟāϰāĻ¤ā§āϰ⧟ āϏāĻŽāϤāĻ˛ā§€ā§Ÿ?

 

(A)  5

(B)  4

(C)  3

(D)  2

 

Answer: (A)  5

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

   $\left|\begin{array}{ccc}2 & 1 & -1 \\ 3 & -2 & 4 \\ 1 & -3 & a\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{ccc}0 & 0 & -1 \\ 11 & 2 & 4 \\ 2 a+1 & a-3 & a\end{array}\right|=0$

$\Rightarrow-(11 a-33-4 a-2)=0$

$\Rightarrow 7 a-35=0$

$\therefore a=5$

Answer: (A) 5

 

15.  $8+4 \sqrt{5} i$ āĻāϰ āĻŦāĻ°ā§āĻ—āĻŽā§‚āϞ āĻšāĻŦ⧇

 

$(\mathrm{A}) \pm(3-2 \mathrm{i})$

(B) $\pm \sqrt{10}+\sqrt{2} \mathrm{i}$

(C) $\pm \sqrt{10}-\sqrt{2} \mathrm{i}$

(D) $\pm(3+2 i)$

Answer: (B) $\pm \sqrt{10}+\sqrt{2} \mathrm{i}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$a=8$

$b=4 \sqrt{5}$

$(8+4 \sqrt{5} i)^{\frac{1}{2}}$

$=\pm\left[\left\{\frac{1}{2}\left(\sqrt{a^{2}+b^{2}}+a\right)\right\}^{\frac{1}{2}}+i\left\{\frac{1}{2}\left(\sqrt{a^{2}+b^{2}}-a\right)\right\}^{\frac{1}{2}}\right]$

$=\pm\left\{\frac{1}{2}(8+\sqrt{64+80})\right\}^{\frac{1}{2}}+i\left\{\frac{1}{2}(-8+\sqrt{64+80})\right\}^{\frac{1}{2}}$

$=\pm\left\{\frac{1}{2}(8+12)\right\}^{\frac{1}{2}}+\mathrm{i}\left\{\frac{1}{2}(-8+12)\right\}^{\frac{1}{2}}$

$=\pm \sqrt{10}+\sqrt{2} \mathrm{i}$

Answer: (B) $\pm \sqrt{10}+\sqrt{2} \mathrm{i}$

 

16.  āϝāĻĻāĻŋ $x^{2}+3 x y+5 y^{2}=1$ āĻšā§Ÿ, āϤāĻžāĻšāϞ⧇ $\frac{d y}{d x}$ āϏāĻŽāĻžāύ āĻšāĻŦ⧇

(A) $-\frac{2 x+3 y}{3 x+10 y}$

(B) $\frac{2 x+3 y}{3 x+10 y}$

(C) $\frac{2 x-3 y}{3 x+10 y}$

(D) $\frac{2 x+3 y}{3 x-10 y}$

Answer: (A) $-\frac{2 x+3 y}{3 x+10 y}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$\mathrm{x}^{2}+3 \mathrm{xy}+5 \mathrm{y}^{2}=1$

$\Rightarrow 2 \mathrm{x}+3 \mathrm{y}+3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+10 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\Rightarrow \frac{d y}{d x}(3 x+10 y)=-(2 x+3 y)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2 \mathrm{x}+3 \mathrm{y}}{3 \mathrm{x}+10 \mathrm{y}}$

Answer: $(\mathrm{A})-\frac{2 \mathrm{x}+3 \mathrm{y}}{3 \mathrm{x}+10 \mathrm{y}}$

 

 

17.  $\frac{1}{2}+\frac{1}{3^{2}}+\frac{1}{2^{8}}+\frac{1}{3^{4}}+\frac{1}{2^{5}}+\frac{1}{3^{6}}+\cdots$ āϧāĻžāϰāĻžāϰ āϏāĻŽāĻˇā§āϟāĻŋ āĻšāĻŦ⧇

(A) $\frac{24}{19}$

(B) $\frac{19}{24}$

(C) $\frac{5}{24}$

(D) $\frac{5}{19}$

Answer: (B) $\frac{19}{24}$

18. $x^{2}-5 x-3=0$ āϏāĻŽā§€āĻ•āϰāϪ⧇āϰ āĻŽā§‚āϞāĻĻā§āĻŦ⧟ $Îą$, $β$ āĻšāϞ⧇ $\frac{1}{\alpha}, \frac{1}{\beta}$ āĻŽā§‚āϞāĻŦāĻŋāĻļāĻŋāĻˇā§āϟ āϏāĻŽā§€āĻ•āϰāĻŖāϟāĻŋ āĻšāĻŦ⧇?

(A) $3 x^{2}-5 x+1=0$

(B) $x^{2}+5 x+3=0$

(C) $5 x^{2}-3 x-1=0$

(D) $3 \mathrm{x}^{2}+5 \mathrm{x}-1=0$

Answer: (D) $3 x^{2}+5 x-1=0$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$a x^{2}+b x+c=0$ āϏāĻŽā§€āĻ•āϰāϪ⧇āϰ āĻŽā§‚āϞāĻĻā§āĻŦ⧟ $alpha$ āĻ“ $beta$ āĻšāĻ˛ā§‡Â $\frac{1}{\alpha}$ āĻ“Â $\frac{1}{\beta}$ āĻŽā§‚āϞāĻŦāĻŋāĻļāĻŋāĻˇā§āϟ āϏāĻŽā§€āĻ•āϰāĻŖ āĻšāĻŦ⧇, $c x^{2}+b x+a=0$

Answer: (D) 3x2 + 5x ‒ 1 = 0

 

19.  āĻŦāĻžāĻ¸ā§āϤāĻŦ āϏāĻ‚āĻ–ā§āϝāĻžā§ŸÂ  $\frac{1}{|2 x-3|}>5$  āĻ…āϏāĻŽāϤāĻžāϟāĻŋāϰ āϏāĻŽāĻžāϧāĻžāύ āĻšāϞ⧋

(A) $\left(\frac{7}{5}, \frac{8}{5}\right)$

(B) $\left[\frac{7}{5}, \frac{8}{5}\right]$

(C) $\left(\frac{7}{5}, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \frac{8}{5}\right)$

(D) $\left[\frac{7}{5}, \frac{3}{2}\right] \cup\left[\frac{3}{2}, \frac{8}{5}\right]$

Answer: (C) $\left(\frac{7}{5}, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \frac{8}{5}\right)$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$|2 x-3|<\frac{1}{5}$

$\Rightarrow-\frac{1}{5}<2 x-3<\frac{1}{5}$

$\Rightarrow \frac{7}{5}<x<\frac{8}{5}$

āφāĻŦāĻžāϰ, $x=\frac{3}{2}$ āĻšāϞ⧇ $\frac{1}{|2 x-3|}=\frac{1}{0}$ āĻ…āϏāĻ‚āĻœā§āĻžāĻžā§ŸāĻŋāϤāĨ¤

Answer: (C) $\left(\frac{7}{5}, \frac{3}{2}\right) \cup\left(\frac{3}{2}, \frac{8}{5}\right)$

 

 

20.  āĻŽāĻžāύ āύāĻŋāĻ°ā§āϪ⧟ āĻ•āϰ: $\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}$

 

(A) 1
(B) $-1$
(C) 2
(D) 3
Answer: (A) 1

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

l’Hôpital’s rule āĻŦā§āϝāĻŦāĻšāĻžāϰ āĻ•āϰ⧇ āĻĒāĻžāχ,

$\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=\lim _{x \rightarrow 0} e^{x}=1$

Answer: (A) 1

 

 

21.  $\int \sqrt{\frac{1+x}{1-x}} d x=f(x)+c$ āĻšāϞ⧇, $\mathrm{f}(\mathrm{x})$ āĻāϰ āĻŽāĻžāύ

(A) $\sin ^{-1} x+\sqrt{1-x^{2}}$

(B) $\sin ^{-1} x-\sqrt{1-x^{2}}$

(C) $\cos ^{-1} \mathrm{x}-\sqrt{1-\mathrm{x}^{2}}$

(D) $\sin ^{-1} x-\sqrt{1+x^{2}}$

Answer: (B) $\sin ^{-1} \mathrm{x}-\sqrt{1-\mathrm{x}^{2}}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\int \sqrt{\frac{1+x}{1-x}} d x$

$=\int \frac{1+x}{\sqrt{1-x^{2}}} d x$

$=\int \frac{d x}{\sqrt{1-x^{2}}}+=\int \frac{x d x}{\sqrt{1-x^{2}}}$

$=\sin ^{-1} x-\sqrt{1-x^{2}}$

Answer: (B) $\sin ^{-1} \mathrm{x}-\sqrt{1-\mathrm{x}^{2}}$

 

22.  $x^{2}+4 x+2 y=0$ āĻĒāϰāĻžāĻŦ⧃āĻ¤ā§āϤ⧇āϰ āĻļā§€āĻ°ā§āώāĻŦāĻŋāĻ¨ā§āĻĻ⧁ āĻšāĻŦ⧇

(A) $(2,-2)$

(B) $(-2,-2)$

(C) $(-2,2)$

(D) $(2,2)$

Answer: (C) $(-2,2)$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{x}^{2}+4 \mathrm{x}+2 \mathrm{y}=0$

$\Rightarrow(\mathrm{x}+2)^{2}=-2(\mathrm{y}-2) \ldots \ldots \ldots(\mathrm{i})$

āϧāϰāĻŋ,

$\mathrm{x}+2=\mathrm{X}, \mathrm{y}-2=\mathrm{Y}$

$(\mathrm{i}) \Rightarrow \mathrm{X}^{2}=-2 \mathrm{Y}$

āĻļā§€āĻ°ā§āώāĻŦāĻŋāĻ¨ā§āĻĻ⧁āϰ āĻ¸ā§āĻĨāĻžāύāĻžāĻ™ā§āĻ•: $\mathrm{X}=0, \mathrm{Y}=\mathrm{O}$

$\therefore \mathrm{x}=-2, \mathrm{y}=2$

Answer: (C) $(-2,2)$

 

23. $f(x)=4-(x-3)^{2}$ āĻĢāĻžāĻ‚āĻļāύ⧇āϰ āĻĄā§‹āĻŽā§‡āχāύ āĻāĻŦāĻ‚ āϰ⧇āχāĻžā§āϜ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡

 

(A) $\mathbf{R}, \mathbf{R}$

(B) $\mathbf{R}, \mathrm{x} \leq 4$

(C) $x \geq 4, \mathbf{R}$

(D) $\mathbf{R}, x \geq 3$

Answer: (B) $\mathbf{R}, x \leq 4$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

$\mathrm{x}$ āĻāϰ āϏāĻŦ āĻŦāĻžāĻ¸ā§āϤāĻŦ āĻŽāĻžāύ⧇āϰ āϜāĻ¨ā§āϝ $f(x)$ āĻāϰ āĻŦāĻžāĻ¸ā§āϤāĻŦ āĻŽāĻžāύ āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§ŸāĨ¤

∴ āĻĢāĻžāĻ‚āĻļāύ⧇āϰ āĻĄā§‹āĻŽā§‡āχāύ = R

āϧāϰāĻŋ,

$y=f(x)=4-(x-3)^{2}$

$\Rightarrow(x-3)^{2}=4-y$

$\Rightarrow(x-3)=\pm \sqrt{4-y}$

$\Rightarrow x=3 \pm \sqrt{4-y}$

$\mathrm{y} \geq 4$ āĻšāĻ˛ā§‡Â $\mathrm{x}$ āĻāϰ āϜāϟāĻŋāϞ āĻŽāĻžāύ āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ āϝāĻž āĻĄā§‹āĻŽā§‡āχāύ āĻāϰ āϏāĻĻāĻ¸ā§āϝ āύ⧟āĨ¤

∴ $\mathrm{x}$ āĻāϰ āϏāĻŦ āĻŦāĻžāĻ¸ā§āϤāĻŦ āĻŽāĻžāύ⧇āϰ āϜāĻ¨ā§āĻ¯Â $y \leq 4$ āĻšāĻŦ⧇āĨ¤

∴ āĻĢāĻžāĻ‚āĻļāύ⧇āϰ āϰ⧇āχāĻžā§āϜ = $\{x: x \in \mathbf{R}$ āĻāĻŦāĻ‚Â $x \leq 4\}$

Answer: (B) $\mathbf{R}, x \leq 4$

 

24. $\mathrm{x}$-āĻ…āĻ•ā§āώāϕ⧇ $(4, 0)$ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϤ⧇ āĻ¸ā§āĻĒāĻ°ā§āĻļ āĻ•āϰ⧇ āĻāĻŦāĻ‚ āϕ⧇āĻ¨ā§āĻĻā§āĻ°Â $5 x-7 y+1=0$ āϏāϰāϞāϰ⧇āĻ–āĻžāϰ āωāĻĒāϰ āĻ…āĻŦāĻ¸ā§āĻĨāĻŋāϤ āĻāĻŽāύ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āϏāĻŽā§€āĻ•āϰāĻŖ āĻšāĻŦ⧇

(A) $x^{2}+y^{2}-8 x-6 y+9=0$

(B) $x^{2}+y^{2}-8 x+6 y+16=0$

(C) $x^{2}+y^{2}-8 x+6 y+9=0$

(D) $x^{2}+y^{2}-8 x-6 y+16=0$

Answer: (B) $x^{2}+y^{2}-8 x+6 y+16=0$

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

āϧāϰāĻŋ, āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āϏāĻŽā§€āĻ•āϰāĻŖ, $x^{2}+y^{2}+2 g x+2 f y+c=0$

âˆĩ āĻŦ⧃āĻ¤ā§āϤāϟāĻŋ $\mathrm{x}$ āĻ…āĻ•ā§āώāϕ⧇ $(4,0)$ āĻŦāĻŋāĻ¨ā§āĻĻ⧁āϤ⧇ āĻ¸ā§āĻĒāĻ°ā§āĻļ āĻ•āϰ⧇ ∴ āϕ⧇āĻ¨ā§āĻĻā§āϰ⧇āϰ āϭ⧁āϜ $=4=-g$ āĻāĻŦāĻ‚ $c=g^{2}=16$

âˆĩ āϕ⧇āĻ¨ā§āĻĻā§āĻ°Â $5 x-7 y+1=0$ āϰ⧇āĻ–āĻžāϰ āωāĻĒāϰ āĻ…āĻŦāĻ¸ā§āĻĨāĻŋāĻ¤Â $\therefore 5(4)-7(-f)+1=0 \Rightarrow f=3$

∴ āĻŦ⧃āĻ¤ā§āϤ⧇āϰ āϏāĻŽā§€āĻ•āϰāĻŖ, $x^{2}+y^{2}+2 g x+2 f y+c=0=x^{2}+y^{2}-8 x+6 y+16=0$

Answer: (B) $x^{2}+y^{2}-8 x+6 y+16=0$

 

25. $A=\left[\begin{array}{cc}1 & \mathrm{i} \\ -\mathrm{i} & 1\end{array}\right], B=\left[\begin{array}{cc}\mathrm{i} & -1 \\ -1 & -\mathrm{i}\end{array}\right]$ āĻāĻŦāĻ‚ $\mathrm{i}=\sqrt{-1}$ āĻšāϞ⧇ $\mathrm{AB}$ āĻāϰ āĻŽāĻžāύ āĻšāĻŦ⧇

(A) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

(B) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

(C) $\left[\begin{array}{ll}\mathrm{i} & 0 \\ 0 & \mathrm{i}\end{array}\right]$

(D) $\left[\begin{array}{ll}\mathrm{i} & 1 \\ 1 & \mathrm{i}\end{array}\right]$

Answer: (B) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\mathrm{AB}$

$=\left[\begin{array}{cc}1 & \mathrm{i} \\ -\mathrm{i} & 1\end{array}\right] \times\left[\begin{array}{cc}\mathrm{i} & -1 \\ -1 & -\mathrm{i}\end{array}\right]$

$=\left[\begin{array}{cc}\mathrm{i}-\mathrm{i} & -1-\mathrm{i}^{2} \\ -\mathrm{i}^{2}-1 & \mathrm{i}-\mathrm{i}\end{array}\right]$

$=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

Answer: (B) $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

 

26. āĻ¸ā§āĻŦāϰāĻŦāĻ°ā§āĻŖāϗ⧁āϞ⧋āϕ⧇ āϏāĻŦ āϏāĻŽā§Ÿ āĻāĻ•āĻ¤ā§āϰ⧇ āϰ⧇āϖ⧇ KACHUA āĻļāĻŦā§āĻĻāϟāĻŋāϰ āĻŦāĻ°ā§āĻŖāϗ⧁āϞ⧋āϕ⧇ āϏāĻžāϜāĻžāύ⧋āϰ āϏāĻ‚āĻ–ā§āϝāĻž āĻšāĻŦ⧇

 

(A)  24

(B)  72

(C)  144

(D)  8

 

Answer:  (B)  72

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

 

KACHUA āĻļāĻŦā§āĻĻāϟāĻŋāϤ⧇ āĻŽā§‹āϟ āĻŦāĻ°ā§āĻŖ $6$āϟāĻŋ āϝāĻžāϰ $3$āϟāĻŋ āĻ¸ā§āĻŦāϰāĻŦāĻ°ā§āĻŖāĨ¤ āĻ¸ā§āĻŦāϰāĻŦāĻ°ā§āĻŖ 3āϟāĻŋ āϕ⧇ āĻāĻ•āϟāĻŋ āĻŦāĻ°ā§āĻŖ āϧāϰāϞ⧇ āĻŽā§‹āϟ āĻŦāĻ°ā§āĻŖ āĻšā§Ÿ $(6 ‒ 3 + 1) = 4$ āϟāĻŋ āϝāĻžāĻĻ⧇āϰāĻ•ā§‡Â $4 !$ āωāĻĒāĻžā§Ÿā§‡ āϏāĻžāϜāĻžāύ⧋ āϝāĻžā§ŸāĨ¤ āφāϰāĻžāϰ āĻ¸ā§āĻŦāϰāĻŦāĻ°ā§āĻŖ 3āϟāĻŋ āϕ⧇ āύāĻŋāĻœā§‡āĻĻ⧇āϰ āĻŽāĻ§ā§āϝ⧇ $\frac{3 !}{2 !}$ āωāĻĒāĻžā§Ÿā§‡ āϏāĻžāϜāĻžāύ⧋ āϝāĻžā§ŸāĨ¤

∴ āĻŽā§‹āϟ āĻŦāĻŋāĻ¨ā§āϝāĻžāϏ āϏāĻ‚āĻ–ā§āϝāĻžÂ $=4 ! \times \frac{3 !}{2 !}=72$

Answer: (B) 72

 

27. āĻāĻ•āϜāύ āϞ⧋āϕ⧇āϰ $3$ āĻœā§‹ā§œāĻž āĻ•āĻžāϞ⧋ āĻŽā§‹āϜāĻž āĻāĻŦāĻ‚ $2$ āĻœā§‹ā§œāĻž āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻŽā§‹āϜāĻž āφāϛ⧇āĨ¤ āĻāĻ•āĻĻāĻŋāύ āĻ…āĻ¨ā§āϧāĻ•āĻžāϰ⧇ āϤāĻžā§œāĻžāĻšā§ā§œāĻž āĻ•āϰ⧇ āϞ⧋āĻ•āϟāĻŋ āĻ•āĻžāĻĒ⧜ āĻĒāϰāϞāĨ¤ āϏ⧇ āĻĒā§āϰāĻĨāĻŽā§‡ āĻāĻ•āϟāĻŋ āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻŽā§‹āϜāĻž āĻĒāϰāĻžāϰ āĻĒāϰ āĻĒāϰāĻŦāĻ°ā§āϤ⧀ āĻŽā§‹āϜāĻžāĻ“ āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻšāĻ“ā§ŸāĻžāϰ āϏāĻŽā§āĻ­āĻžāĻŦāύāĻž

(A) $\frac{1}{3}$

(B) $\frac{2}{15}$

(C) $\frac{1}{10}$

(D) $\frac{3}{10}$

Answer: (C) $\frac{2}{15}$

 

āϏāĻžāĻŽāĻžāϧāĻžāύ :

                    āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻŽā§‹āϜāĻžāϰ āϏāĻ‚āĻ–ā§āϝāĻž = 2×2 = 4

                    āĻ•āĻžāϞ⧇āĻž āĻŽā§‹āϜāĻžāϰ āϏāĻ‚āĻ–ā§āϝāĻž = 2×3 = 6

āĻĒā§āϰāĻĨāĻŽā§‡ āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻŽā§‹āϜāĻž āĻĒāϰāĻžāϰ āϏāĻŽā§āĻ­āĻŦā§āϝāϤāĻžÂ $=\frac{4}{10}=\frac{2}{5}$

āĻĒāϰ⧇ āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻŽā§‹āϜāĻž āĻĒāϰāĻžāϰ āϏāĻŽā§āĻ­āĻŦā§āϝāϤāĻžÂ $=\frac{3}{9}=\frac{1}{3}$

āĻĒā§āϰāĻĨāĻŽā§‡ āĻāĻ•āϟāĻŋ āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻŽā§‹āϜāĻž āĻĒāϰāĻžāϰ āĻĒāϰ āĻĒāϰāĻŦāĻ°ā§āϤ⧀ āĻŽā§‹āϜāĻžāĻ“ āĻŦāĻžāĻĻāĻžāĻŽā§€ āĻšāĻ“ā§ŸāĻžāϰ āϏāĻŽā§āĻ­āĻžāĻŦāύāĻžÂ $\frac{2}{5} \times \frac{1}{3}=\frac{2}{15}$

Answer: (C) $\frac{2}{15}$

 

28. $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-3}{2 \mathrm{x}+1}$ āĻāĻŦāĻ‚ $\mathrm{x} \neq \frac{1}{2}$ āĻšāϞ⧇ $\mathrm{f}^{-1}(-2)$ āĻāϰ āĻŽāĻžāύ āĻ•āϤ āĻšāĻŦ⧇

(A) $\frac{1}{2}$

(B) $\frac{1}{5}$

(C) 2

(D) 5

Answer: (B) $\frac{1}{5}$

āϏāĻžāĻŽāĻžāϧāĻžāύ :

āϧāϰāĻŋ,

$\mathrm{y}=\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-3}{2 \mathrm{x}+1}$

$\Rightarrow 2 \mathrm{xy}+\mathrm{y}=\mathrm{x}-3$

$\Rightarrow 2 \mathrm{xy}-\mathrm{x}=-\mathrm{y}-3$

$\Rightarrow \mathrm{x}=\frac{\mathrm{y}+3}{1-2 \mathrm{y}}$

$\mathrm{f}^{-1}(\mathrm{x})=\frac{\mathrm{x}+3}{1-2 \mathrm{x}}$

$f^{-1}(-2)=\frac{-2+3}{1-2(-2)}=\frac{1}{5}$

Answer: (B) $\frac{1}{5}$

29.  $\mathrm{u}$ āĻŦ⧇āϗ⧇ āĻ…āύ⧁āϭ⧁āĻŽāĻŋāϕ⧇āϰ āϏāĻžāĻĨ⧇ $alpha$ āϕ⧋āϪ⧇ āĻĒā§āϰāĻ•ā§āώāĻŋāĻĒā§āϤ āĻŦāĻ¸ā§āϤ⧁āϰ āϏāĻ°ā§āĻŦā§‹āĻšā§āϚ āωāĻšā§āϚāϤāĻž āĻšāĻŦ⧇

(A) $\frac{\mathrm{u}^{2} \sin 2 \alpha}{2 \mathrm{~g}}$

(B) $\frac{\mathrm{u}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}$

(C) $\frac{\mathrm{u}^{2} \sin 2 \alpha}{\mathrm{g}}$

(D) $\frac{\mathrm{u}^{2} \sin ^{2} \alpha}{\mathrm{g}}$

Answer: (B) $\frac{\mathrm{u}^{2} \sin ^{2} \alpha}{2 \mathrm{~g}}$

 

 

30. $\frac{(i+1)^{2}}{(i-1)^{4}}$ āϜāϟāĻŋāϞ āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋāϰ āφāĻ°ā§āϗ⧁āĻŽā§‡āĻ¨ā§āϟ āĻšāĻŦ⧇

(A)  Ī€

(B)  ‒ Ī€

(C)  $\frac{\pi}{2}$

(D)  ‒ $\frac{\pi}{2}$

 

Answer: (D)  ‒ $\frac{\pi}{2}$

āϏāĻžāĻŽāĻžāϧāĻžāύ :

$\frac{(i+1)^{2}}{(i-1)^{4}}$

$=\frac{(i+1)^{2}}{(i-1)^{2}(i-1)^{2}}$

$=\frac{\mathrm{i}^{2}+2 \mathrm{i}+1}{\left(\mathrm{i}^{2}-2 \mathrm{i}+1\right)\left(\mathrm{i}^{2}-2 \mathrm{i}+1\right)}$

$=\frac{2 \mathrm{i}}{(-2 \mathrm{i})(-2 \mathrm{i})}$

$=\frac{2 \mathrm{i}}{4 \mathrm{i}^{2}}$

$=-\frac{1}{2} \mathrm{i}$

∴ āϜāϟāĻŋāϞ āϏāĻ‚āĻ–ā§āϝāĻžāϟāĻŋāϰ āφāĻ°ā§āϗ⧁āĻŽā§‡āĻ¨ā§āĻŸÂ $-\frac{\pi}{2}$

Answer: (D)  ‒ $\frac{\pi}{2}$

 

āϰāϏāĻžā§Ÿāύ

 

 

1. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āĻ•ā§‹ā§ŸāĻžāĻ¨ā§āϟāĻžāĻŽ āϏ⧇āϟāϟāĻŋ āĻĒāϰāĻŽāĻžāύ⧁āϰ āĻāĻ•āϟāĻŋ āχāϞ⧇āĻ•āĻŸā§āϰāύ⧇āϰ āϜāĻ¨ā§āϝ āϏāĻŽā§āĻ­āĻŦ āύ⧟ ?

(A) $\mathrm{n}=2, \mathrm{I}=1, \mathrm{~m}=\mathrm{o}, \mathrm{s}=+1 / 2$

(B) $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=2, \mathrm{~s}=-1 / 2$

(C) $\mathrm{n}=1, \mathrm{I}=0, \mathrm{~m}=0, \mathrm{~s}=-1 / 2$

(D) $n=2, \mathrm{I}=0, \mathrm{~m}=0, \mathrm{~s}=+1 / 2$

 

āωāĻ¤ā§āϤāϰ : (B) $\mathrm{n}=3, \mathrm{l}=1, \mathrm{~m}=2, \mathrm{~s}=-1 / 2$

 

2. āĻšāĻžāχāĻĄā§āϰ⧋āĻœā§‡āύ⧇āϰ āĻĒāĻžāϰāĻŽāĻžāĻŖāĻŦāĻŋāĻ• āĻŦāĻ°ā§āĻŖāĻžāϞ⧀āϰ āϕ⧋āύ āϏāĻŋāϰāĻŋāϜāϟāĻŋāϤ⧇ āĻĻ⧃āĻļā§āϝāĻŽāĻžāύ āĻ…āĻžā§āϚāϞ⧇āϰ āϰāĻļā§āĻŽāĻŋ āĻĻ⧇āĻ–āĻž āϝāĻžā§Ÿ ?

 

(A) Paschen

(B)  Lyman

(C)  Balmer

(D)  Brackett

 

āωāĻ¤ā§āϤāϰ : (C) Balmer

 

3. $10.0g$ āĻ…āĻ•ā§āϏāĻŋāĻœā§‡āύ⧇ āĻ…āϪ⧁āϰ āϏāĻ‚āĻ–ā§āϝāĻž āĻ•āϤ ?

 

(A) $3.76 \times 10^{23}$

(B) $6.02 \times 10^{22}$

(C) $9.63 \times 10^{23}$

(D) $1.88 \times 10^{23}$

 

āωāĻ¤ā§āϤāϰ : (D) $1.88 \times 10^{23}$

 

4. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύāϟāĻŋāϰ āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ āϏāĻŦāĻšā§‡ā§Ÿā§‡ āĻ•āĻŽ ?

 

(A) $_{9} \mathrm{~F}^{-}$

(B) ${ }_{10} \mathrm{Ne}$

(C) ${ }_{11} \mathrm{Na}^{+}$

(D) $12 \mathrm{Mg}^{2+}$

 

āωāĻ¤ā§āϤāϰ : (A) $_{9} \mathrm{~F}^{-}$

 

5. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āϝ⧌āĻ—āϟāĻŋ āϏāĻŦāĻšā§‡ā§Ÿā§‡ āĻ•āĻŽ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžā§Ÿ āĻŦāĻŋā§Ÿā§‹āϜāĻŋāϤ āĻšāĻŦ⧇ ?

 

(A) $\mathrm{Na}_{2} \mathrm{CO}_{3}$

(B) $\mathrm{K}_{2} \mathrm{CO}_{3}$

(C) $\mathrm{MgCO}_{3}$

(D) $\mathrm{BaCO}_{3}$

 

āωāĻ¤ā§āϤāϰ : (C) $\mathrm{MgCO}_{3}$

 

6.āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āϝ⧌āĻ—āϟāĻŋāϰ āφāĻ•ā§ƒāϤāĻŋ āĻ¤ā§āϰāĻŋāϕ⧋āĻŖāĻžāĻ•āĻžāϰ āϏāĻŽāϤāĻ˛ā§€ā§Ÿ ?

 

(A) $\mathrm{BCl}_{3}$

(B) $\mathrm{H}_{3} \mathrm{O}^{+}$

(C) $\mathrm{BrF}_{5}$

(D) $\mathrm{PH}_{3}$

 

āωāĻ¤ā§āϤāϰ : (A) $\mathrm{BCl}_{3}$

 

7. $300 k$ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžā§Ÿ āĻŦāĻžāϤāĻžāϏ⧇āϰ $\mathrm{N}_{2}$ āĻ…āϪ⧁āϰ āĻ—āϤāĻŋ āĻ•āϤ ?

(A) $450 \mathrm{~m} / \mathrm{s}$

(B) $516 \mathrm{~m} / \mathrm{s}$

(C) $400 \mathrm{~m} / \mathrm{s}$

(D) $600 \mathrm{~m} / \mathrm{s}$

 

āωāĻ¤ā§āϤāϰ : (B) $516 \mathrm{~m} / \mathrm{s}$

 

8. āĻ—ā§āϰāĻŋāĻ—āύāĻžāϰāĻĄ āĻŦāĻŋāĻ•āĻžāϰāĻ• āĻšāϞ⧋ –

(A) CH3ONa

(B) RBaCl

(C) RMgX

(D) RCaX

 

 

āωāĻ¤ā§āϤāϰ : (C) RMgX

 

9. āχāĻĨāĻžāύāϞ⧇āĻ°Â $170^{\circ} \mathrm{C}$ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžā§Ÿ āĻ…āϤāĻŋāϰāĻŋāĻ•ā§āϤ āĻ—āĻžā§ āϏāĻžāϞāĻĢāĻŋāωāϰāĻŋāĻ• āĻāϏāĻŋāĻĄ āĻĻā§āĻŦāĻžāϰāĻž āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻž āĻ•āϰāĻžāϞ⧇ āϕ⧀ āĻ‰ā§ŽāĻĒāĻ¨ā§āύ āĻšā§Ÿ ?

(A) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SO}_{4}$

(B) $\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{SO}_{4}$

(C) $\mathrm{CH}_{2}=\mathrm{CH}_{2}$

(D) $\mathrm{CH}_{3} \mathrm{CHO}$

 

āωāĻ¤ā§āϤāϰ : (C) $\mathrm{CH}_{2}=\mathrm{CH}_{2}$

10. $10.0$ āĻŽāĻŋ.āϞāĻŋ. $\mathrm{NaOH}$ āĻĻā§āϰāĻŦāĻŖāϕ⧇ $0.12 \mathrm{M}$ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻžā§Ÿ $15.0$ āĻŽāĻŋ.āϞāĻŋ āĻ…āĻ•ā§āϏāĻžāϞāĻŋāĻ• āĻāϏāĻŋāĻĄ āĻĻā§āĻŦāĻžāϰāĻž āϟāĻžāχāĻŸā§āϰ⧇āĻļāύ āĻ•āϰāϞ⧇ āĻĒā§āϰāĻļāĻŽāύ āĻŦāĻŋāĻ¨ā§āĻĻ⧁ āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ āĨ¤Â $\mathrm{NaOH}$ āĻāϰ āϘāύāĻŽāĻžāĻ¤ā§āϰāĻž āĻ•āϤ ?

(A) $0.25 \mathrm{M}$

(B) $0.36 \mathrm{M}$

(C) $0.32 \mathrm{M}$

(D) $0.40 \mathrm{M}$

 

āωāĻ¤ā§āϤāϰ : (B) $0.36 \mathrm{M}$

 

11.āĻŽā§āϝāĻžāύ⧇āϏāĻŋ⧟āĻžāĻŽ āĻĢāϏāĻĢ⧇āĻŸā§‡āϰ āϏāĻ‚āϕ⧇āϤ āĻšāϞ⧋ -

 

(A) $\mathrm{Mg}_{2}\left(\mathrm{PO}_{4}\right)_{2}$

(B) $\mathrm{MgPO}_{4}$

(C) $\mathrm{Mg}_{2}\left(\mathrm{PO}_{4}\right)_{2}$

(D) $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$

 

āωāĻ¤ā§āϤāϰ : (D) $\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}$

 

12. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžāϟāĻŋ āĻĄāĻŋāϏāĻĒā§āϰ⧋āĻĒāϰāύ⧇āĻļāύ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻž ?

 

(A) $\mathrm{H}_{2} \mathrm{~S}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}+\mathrm{S}$

(B) $\mathrm{CuSO}_{4}+\mathrm{NH}_{4} \mathrm{OH} \rightarrow \mathrm{Cu}(\mathrm{OH})_{2}+\left(\mathrm{NH}_{3}\right)_{2} \mathrm{SO}_{4}$

(C) $\mathrm{Fe}+$ dil. $\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{FeSO}_{4}+\mathrm{H}_{2}$

(D) $\mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$

 

āωāĻ¤ā§āϤāϰ : (D) $\mathrm{Cl}_{2}+\mathrm{NaOH} \rightarrow \mathrm{NaOCl}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}$

 

13.$18.5 \% \mathrm{~N}_{2} \mathrm{O}_{4}$ $25^{\circ} \mathrm{C}$ āϤāĻžāĻĒāĻŽāĻžāĻ¤ā§āϰāĻžā§Ÿ āĻāĻŦāĻ‚Â $1 \mathrm{~atm}$ āϚāĻžāĻĒ⧇ āĻŦāĻŋā§Ÿā§‹āϜāĻŋāϤ āĻšāϞ⧇ $K_{p}$ āĻāϰ āĻŽāĻžāύ āĻ•āϤ ?

 

(A) $0.142 \mathrm{~atm}$

(B) $0.185 \mathrm{~atm}$

(C) $0.220 \mathrm{~atm}$

(D) $0.125 \mathrm{~atm}$

 

āωāĻ¤ā§āϤāϰ : (A) $0.142 \mathrm{~atm}$

 

 

14. āύāĻŋāĻŽā§āύ⧇āϰ āĻŽāĻŋāĻļā§āϰāĻŖāϏāĻŽā§‚āĻš āĻĨ⧇āϕ⧇ āĻŦāĻžāĻĢāĻžāϰ āĻĻā§āϰāĻŦāĻŖāϟāĻŋ āĻļāύāĻžāĻ•ā§āϤ āĻ•āϰ -

 

(A) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$

(B) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.1 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$

(C) $0.1 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$

(D) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{HCl}+0.1 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$

 

āωāĻ¤ā§āϤāϰ : (A) $0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{CH}_{3} \mathrm{COOH}+0.2 \mathrm{M}$ $10 \mathrm{~mL}$ $\mathrm{NaOH}$

 

 

15. āύāĻŋāĻšā§‡āϰ āϕ⧋āύ āϝ⧌āĻ—āϟāĻŋ Fehling āĻĻā§āϰāĻŦāϪ⧇āϰ āϏāĻžāĻĨ⧇ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻž āĻ•āϰ⧇ āϞāĻžāϞ āĻ…āϧāσāĻ•ā§āώ⧇āĻĒ āĻĻā§‡ā§Ÿ ?

 

(A) $\mathrm{RCH}_{2} \mathrm{X}$

(B) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}$

(C) $\mathrm{RCH}_{2} \mathrm{OH}$

(D) $\mathrm{RCH}_{2} \mathrm{CHO}$

 

āωāĻ¤ā§āϤāϰ : (D) $\mathrm{RCH}_{2} \mathrm{CHO}$

 

16. āĻāĻ•āϟāĻŋ $\mathrm{CH}_{3} \mathrm{CN}$ āĻ…āϪ⧁āĻ¤ā§‡Â $\pi$ āĻāĻŦāĻ‚Â $\sigma$ āĻŦāĻ¨ā§āϧāύ⧇āϰ āϏāĻ‚āĻ–ā§āϝāĻž āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡ -

(A) 5 and 2

(B) 4 and 3

(C) 5 and 3

(D) 4 and 2

 

āωāĻ¤ā§āϤāϰ : (A) 5 and 2

 

17. $\mathrm{IUPAC}$ āύāĻžāĻŽāĻ•āϰāĻŖ āĻ…āύ⧁āϏāĻžāϰ⧇ $\mathrm{CH}_{3}-\mathrm{CH}\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)-\mathrm{CH}_{2}-\mathrm{CHBr}-\mathrm{CHCl}-\mathrm{CH}_{3}$ āĻāϰ āύāĻžāĻŽ āĻšāϞ⧋ -

 

(A)  2-chloro-3-bromo-5-ethylhexane

(B)  2-chloro-3-bromo-5-methyl heptane

(C)  3-bromo-2-chloro-5-ethyl-hexane

(D)  3-bromo-2-chloro-5-methyl heptane

 

āωāĻ¤ā§āϤāϰ :  (B)  2-chloro-3-bromo-5-methyl heptane

 

18. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύāϟāĻŋ āϏāĻŦāĻšā§‡ā§Ÿā§‡ āĻ¸ā§āĻĨāĻŋāϤāĻŋāĻļā§€āϞ āĻ•āĻžāĻ°ā§āĻŦā§‹-āĻ•ā§āϝāĻžāϟāĻžā§Ÿāύ ?

 

(A) $\mathrm{CH}_{3}{ }^{+}$

(B) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}^{+} \mathrm{H}$

(C) $\mathrm{H}_{2} \mathrm{C}^{+}-\mathrm{CH}_{3}$

(D) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$

 

āωāĻ¤ā§āϤāϰ : (D) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$

 

19. āĻĒāϟāĻžāϏāĻŋ⧟āĻžāĻŽ āϏāĻžā§ŸāĻžāύāĻžāχāĻĄā§‡āϰ āωāĻĒāĻ¸ā§āĻĨāĻŋāϤāĻŋāϤ⧇ āĻšāĻžāχāĻĄā§āϰ⧋āĻœā§‡āύ āϏāĻžā§ŸāĻžāύāĻžāχāĻĄ āĻĒā§āϰāĻžāĻĒāĻžāύāϞ⧇āϰ āϏāĻžāĻĨ⧇ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻž āĻ•āϰ⧇ āĨ¤āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžāϰ āĻ•ā§ŒāĻļāϞāϟāĻŋ āĻšāϞ⧋ -

 

(A)  nuclophilic addition

(B)  electrophilic substitution

(C)  nucleophilic substituton

(D)  electrophilic addition

 

āωāĻ¤ā§āϤāϰ : (C)  nucleophilic substituton

 

 

20. āĻ…ā§āϝāĻžāϰ⧋āĻŽā§‡āϟāĻŋāĻ• āĻĒā§āϰāϤāĻŋāĻ¸ā§āĻĨāĻžāĻĒāύ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžā§Ÿ āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āĻ—ā§āϰ⧁āĻĒāϟāĻŋ āĻŽā§‡āϟāĻž āύāĻŋāĻ°ā§āĻĻ⧇āĻļāĻ• ?

(A) $-\mathrm{NO}_{3}$

(B) $-\mathrm{OH}$

(C) $-\mathrm{Cl}$

(D) $-\mathrm{CH}_{3}$

 

āωāĻ¤ā§āϤāϰ : (A) $-\mathrm{NO}_{3}$

 

21. āϏ⧋āĻĄāĻž āĻ…ā§āϝāĻžāĻļ āĻļāĻŋāĻ˛ā§āĻĒāĻ•ā§āώ⧇āĻ¤ā§āϰ⧇ āϕ⧋āύ āĻĒāĻĻā§āϧāϤāĻŋāϤ⧇ āϤ⧈āϰāĻŋ āĻ•āϰāĻž āĻšā§Ÿ ?

 

(A)  Bessemer process

(B)  Chamber process

(C)  Solvary process

(D)  Haber process

 

āωāĻ¤ā§āϤāϰ : (C) Solvary process

 

22. āĻāĻŦāĻ‚ āĻ…āϧāσāĻ•ā§āώ⧇āĻĒ āϏāĻŽā§‚āĻšā§‡āϰ āϰāĻ‚-āĻāϰ āĻ•ā§āϰāĻŽ āĻšāϞ⧋ –

 

(A)brown, pink, white and blue

(B)brown, blue white and pink

(C)pink, white , brown and blue

(D) brown, white, blue and pink

 

āωāĻ¤ā§āϤāϰ : (B)brown, blue white and pink

 

23. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻ…āϏāĻ¤ā§āϝ ?

(A) $\mathrm{NH}_{4}^{+}$is the conjugate acid of base $\mathrm{NH}_{3}$

(B) $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ are conjugate pair

(C) $\mathrm{OH}$ is the conjugate base of acid $\mathrm{H}_{2} \mathrm{O}$

(D) $\mathrm{OH}^{-}$and $\mathrm{H}_{2} \mathrm{O}$ are conjugate pair

 

āωāĻ¤ā§āϤāϰ : (B) $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{O}$ are conjugate pair

 

24. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āϝ⧌āĻ—āϟāĻŋ āϜāĻ˛ā§€ā§Ÿ āĻĻā§āϰāĻŦāϪ⧇ āϏāĻŦāĻšā§‡ā§Ÿā§‡ āϏāĻšāĻœā§‡ āĻšāĻžāχāĻĄā§āϰ⧋-āĻŦāĻŋāĻļā§āϞ⧇āώāĻŋāϤ āĻšā§Ÿ ?

(A) $\mathrm{CCl}_{4}$

(B) $\mathrm{SnCl}_{2}$

(C) $\mathrm{SiCl}_{4}$

(D) $\mathrm{PbCl}_{4}$

 

āωāĻ¤ā§āϤāϰ : (C) $\mathrm{SiCl}_{4}$

 

25. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āϝ⧌āĻ—āϟāĻŋāϤ⧇ āϏāĻžā§āϚāĻžāϞāύ āĻ…āĻ•ā§āώāĻŽ $Ī€$–āχāϞ⧇āĻ•ā§āĻŸā§āϰāύ āφāĻ› ?

 

(A) $\mathrm{C}_{2} \mathrm{H}_{6}$

(B) $\mathrm{C}_{6} \mathrm{H}_{6}$

(C) $\mathrm{C}_{3} \mathrm{H}_{8}$

(D) $\mathrm{C}_{2} \mathrm{H}_{4}$

 

āωāĻ¤ā§āϤāϰ : (D) $\mathrm{C}_{2} \mathrm{H}_{4}$

 

26. $\mathrm{Sr}, \mathrm{Tc}, \mathrm{Zr}$ āĻāĻŦāĻ‚Â $\mathrm{Rb}$ āĻĒāϰāĻŽāĻžāύ⧁āϰ āĻŦā§āϝāĻžāϏāĻžāĻ°ā§āϧ⧇āϰ āĻ•ā§āϰāĻŽ āĻšāϞ⧋ –

 

(A) $\mathrm{Rb}>\mathrm{Sr}>\mathrm{Zr}>\mathrm{Tc}$

(B) $\mathrm{Tc}>\mathrm{Sr}>\mathrm{Rb}>\mathrm{Zr}$

(C) $\mathrm{Sr}>\mathrm{Tc}>\mathrm{Zr}>\mathrm{Rb}$

(D) $\mathrm{Zr}>\mathrm{Tc}>\mathrm{Rb}>\mathrm{Sr}$

 

āωāĻ¤ā§āϤāϰ : (A) $\mathrm{Rb}>\mathrm{Sr}>\mathrm{Zr}>\mathrm{Tc}$

 

 

27. āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžāϟāĻŋ āĻĒā§āϰāĻļāĻŽāύ āĻāύāĻĨāĻžāϞāĻĒāĻŋ, $\Delta H_{\text {neutr }}$ āĻĒā§āϰāĻ•āĻžāĻļ āĻ•āϰ⧇ ?

 

(A) $\mathrm{H}_{2} \mathrm{SO}_{4}$ (aq) $+\mathrm{Ca}(\mathrm{OH})_{2}$ (aq) $\rightarrow \mathrm{CaSO}_{4}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l

(B) $\mathrm{H}_{2} \mathrm{SO}_{4}$ (aq) $+2 \mathrm{NH}_{3}$ (aq) $\rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}$ (aq)

(C) $\mathrm{HCl}$ (aq) $+1^{1 / 2} \mathrm{Ca}(\mathrm{OH})_{2}$ (aq) $\rightarrow 1 / 2 \mathrm{CaCl}_{2}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l)

(D) $2 \mathrm{HCl}$ (aq) $+\mathrm{Ca}$ (OH) $_{2}$ (aq) $\rightarrow \mathrm{CaCl}_{2}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l)

 

āωāĻ¤ā§āϤāϰ : (C) $\mathrm{HCl}$ (aq) $+1^{1 / 2} \mathrm{Ca}(\mathrm{OH})_{2}$ (aq) $\rightarrow 1 / 2 \mathrm{CaCl}_{2}$ (aq) $+2 \mathrm{H}_{2} \mathrm{O}$ (l)

 

28. $\mathrm{Zn}^{+} \mid \mathrm{Zn}$ āĻāĻŦāĻ‚ $\mathrm{Ag}^{+} \mid \mathrm{Ag}$ āϤ⧜āĻŋā§ŽāĻĻā§āĻŦāĻžāϰ āĻĻ⧁āϟāĻŋāϰ āĻŦāĻŋāϜāĻžāϰāĻŖ āĻŦāĻŋāĻ­āĻŦ āϝāĻĨāĻžāĻ•ā§āϰāĻŽā§‡Â $-0.76 \mathrm{~V}$ āĻāĻŦāĻ‚Â $+0.80 \mathrm{~V}$ āĻāχ āϤ⧜āĻŋā§ŽāĻĻā§āĻŦāĻžāϰ āĻĻ⧁āϟāĻŋ āĻĻā§āĻŦāĻžāϰāĻž āϤ⧈āϰāĻŋ āϕ⧋āώ⧇āϰ āĻŽā§‹āϟ āĻŦāĻŋāĻ­āĻŦ āĻ•āϤ āĻšāĻŦ⧇ ?

 

(A) $-0.04 \mathrm{~V}$

(B) $+1.56 \mathrm{~V}$

(C) +0.14 V

(D) $-1.56 \mathrm{~V}$

 

āωāĻ¤ā§āϤāϰ : (B) $+1.56 \mathrm{~V}$

 

29.āĻŦā§‹āĻ˛ā§āϟāϜāĻŽā§āϝāĻžāύ āĻ§ā§āϰ⧁āĻŦāϕ⧇āϰ āĻāĻ•āĻ• āĻšāϞ⧋ –

(A) J /molecule

(B) J.s

(C) $\mathrm{J} / \mathrm{K}$

(D) g/cc

 

āωāĻ¤ā§āϤāϰ : (A) J /molecule

 

30. $\operatorname{Sn}(s)+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \operatorname{Sn}^{+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ -  āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻžāϟāĻŋāϰ āĻ•ā§āώ⧇āĻ¤ā§āϰ⧇ āύāĻŋāĻŽā§āύ⧇āϰ āϕ⧋āύāϟāĻŋ āϕ⧋āώ⧇āϰ āĻŦ⧃āĻĻā§āϧāĻŋ āĻ•āϰāĻŦ⧇ ?

 

(A) icrease in the size of silver rod

(B) increase in the concentration of Sn+ ions

(C) increase in the concentration of Ag+ ions

(D) increase in the size of tin rod

 

āωāĻ¤ā§āϤāϰ : (C) increase in the concentration of Ag+ ions

 

āĻœā§€āĻŦāĻŦāĻŋāĻœā§āĻžāĻžāύ

 

ā§§. āĻŦāĻžāĻ¸ā§āϟ āϤāĻ¨ā§āĻ¤ā§āϰ āϕ⧋āύāϟāĻŋ ?

       āĻ•. āĻĒāĻžāϟ āϤāĻ¨ā§āĻ¤ā§āϰ

       āĻ–. āĻ•āĻžāĻ°ā§āĻĒāĻžāϏ āϤāĻ¨ā§āĻ¤ā§āϰ

       āĻ—. āĻļāĻŋāĻŽā§āϞ āϤ⧁āϞāĻž

       āϘ. āĻ•ā§Ÿā§‡āϰ

 

āωāĻ¤ā§āϤāϰ : (āĻ•) āĻĒāĻžāϟ āϤāĻ¨ā§āĻ¤ā§āϰ

 

⧍. āĻŽāĻžāύ⧁āώ⧇āϰ āĻŦāĻ•ā§āώāĻĻ⧇āĻļā§€ā§Ÿ āĻ•āĻļ⧇āϰ⧁āĻ•āĻž āĻ•ā§ŸāϟāĻŋ ?

       āĻ•. 7

       āĻ–. 12

       āĻ—. 10

       āϘ. 15

 

āωāĻ¤ā§āϤāϰ : āĻ–. 12

 

ā§Š. āϕ⧋āύāϟāĻŋ ommatidium āĻāϰ āĻ…āĻ‚āĻļ āύ⧟ ?

 

       āĻ•. rhabdosome

       āĻ–. rentinal sheath

       āĻ—. retinal cell

       āϘ. ocellus

 

āωāĻ¤ā§āϤāϰ : āϘ. ocellus

 

ā§Ē. āĻ—ā§‹āĻĻāϰ⧋āĻ— āϏ⧃āĻˇā§āϟāĻŋāĻ•āĻžāϰ⧀ āĻĒāϰāĻœā§€āĻŦāĻŋāϰ āύāĻžāĻŽ-

       āĻ•. entamoeba histolytica

       āĻ–. Wuchereria bancrofti

       āĻ—. Ades fatigans

       āϘ. Culex quinquefasciatus

 

āωāĻ¤ā§āϤāϰ : āĻ–. Wuchereria bancrofti

 

ā§Ģ. Liliopsida āĻŦāϞāϤ⧇ āĻ•āĻŋ āĻŦ⧁āĻāĻžā§Ÿ?

       āĻ•. āĻāĻ•āĻŦā§€āϜāĻĒāĻ¤ā§āϰ⧀ āωāĻĻā§āĻ­āĻŋāĻĻ

       āĻ–. āĻĻā§āĻŦāĻŋāĻŦā§€āϜāĻĒāĻ¤ā§āϰ⧀ āωāĻĻā§āĻ­āĻŋāĻĻ

       āĻ—. āϞāĻŋāϞāĻŋā§Ÿā§‡āϏ⧀ āĻ—ā§‹āĻ¤ā§āϰ

       āϘ. āϞāĻŋāϞāĻŋā§Ÿā§‡āϞāĻŋāϏ āĻŦāĻ°ā§āĻ—

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻāĻ•āĻŦā§€āϜāĻĒāĻ¤ā§āϰ⧀ āωāĻĻā§āĻ­āĻŋāĻĻ

 

ā§Ŧ. āφāϰāĻļā§‹āϞāĻžāϰ āĻĒ⧇āϰāĻŋāĻŸā§āϰāĻĢāĻŋāĻ• āĻĒāĻ°ā§āĻĻāĻž āĻĨāĻžāϕ⧇ āϕ⧋āύ āĻ¸ā§āĻĨāĻžāύ⧇?

       āĻ•. āĻ•ā§āϰāĻĒ

       āĻ–. āĻ—āĻŋāϜāĻžāĻ°ā§āĻĄ

       āĻ—. āĻšā§‡āĻĒāĻžāϟāĻŋāĻ• āϏāĻŋāĻ•āĻžāĻŽ

       āϘ. āĻŽā§‡āϏ⧇āύāϟāĻŋāϰāύ

 

āωāĻ¤ā§āϤāϰ :   āϘ. āĻŽā§‡āϏ⧇āύāϟāĻŋāϰāύ

 

ā§­. F1-āϜāύ⧁āϰ āωāĻĻā§āĻ­āĻŋāĻĻāϕ⧇ āĻĒā§āϰāĻšā§āĻ›āĻ¨ā§āύ āĻĒā§āϝāĻžāϰ⧇āĻ¨ā§āĻŸā§‡āϰ āϏāĻžāĻĨ⧇ āĻ•ā§āϰāϏ āĻ•āϰāĻžāϕ⧇ āĻŦāϞāĻž āĻšā§Ÿ

       āĻ•. āĻŦā§āϝāĻžāĻ• āĻ•ā§āϰāϏ

       āĻ–. āĻŸā§‡āĻ¸ā§āϟ āĻ•ā§āϰāϏ

       āĻ—. āĻŽāύ⧋āĻšāĻžāχāĻŦā§āϰāĻŋāĻĄ āĻ•ā§āϰāϏ

       āϘ. āĻĄāĻžāχāĻšāĻžāχāĻŦā§āϰāĻŋāĻĄ āĻ•ā§āϰāϏ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āĻŸā§‡āĻ¸ā§āϟ āĻ•ā§āϰāϏ

 

ā§Ž. āĻŽāĻžāύ⧁āώ⧇āϰ āĻĻ⧇āĻšā§‡ āϕ⧋āύāϟāĻŋ āĻŸā§‡āĻ¸ā§āĻŸā§‹āĻ¸ā§āĻŸā§‡āϰāύ āϤ⧈āϰāĻŋ āĻ•āϰ⧇?

 

       āĻ•. āĻ¸ā§āĻ•ā§āϰ⧋āϟāĻžāĻŽ

       āĻ–. āχāĻ¨ā§āϟāĻžāϰāϟāĻŋāĻ¸ā§āϟāĻŋāĻļāĻŋ⧟āĻžāϞ āϕ⧋āώ

       āĻ—. āχāĻĒāĻŋāĻĄāĻŋāĻĄāĻžāχāĻŽāĻŋāϏ

       āϘ. āĻ¸ā§āĻĒāĻžāϰāĻŽā§‡āĻŸā§‹āĻ—ā§‹āύāĻŋ⧟āĻž

 

āωāĻ¤ā§āϤāϰ :   āĻ–. āχāĻ¨ā§āϟāĻžāϰāϟāĻŋāĻ¸ā§āϟāĻŋāĻļāĻŋ⧟āĻžāϞ āϕ⧋āώ

 

 

⧝. āĻ•ā§āϰ⧇āĻŦāϏ āϚāĻ•ā§āϰ⧇ āĻ•āϤāϟāĻŋ NADH2 āϤ⧈āϰāĻŋ āĻšā§Ÿ?

       āĻ•. 1

       āĻ–. 2

       āĻ—. 3

       āϘ. 4

 

āωāĻ¤ā§āϤāϰ : āĻ—. 3

 

ā§§ā§Ļ. āϕ⧋āϞāĻžāĻ°ā§Ÿā§‡āĻĄ āĻŽā§‚āϞ āϕ⧋āĻĨāĻžā§Ÿ āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ?

 

       āĻ•. Cycas

       āĻ–. Ficus

       āĻ—. Hibiscus

       āϘ. Daucus

 

āωāĻ¤ā§āϤāϰ : āĻ•. Cycas

 

ā§§ā§§. āχāύāϏ⧁āϞāĻŋāύ āύāĻŋāσāϏāϰāĻŖāĻ•āĻžāϰ⧀ āĻ—ā§āϰāĻ¨ā§āĻĨāĻŋāϰ āύāĻžāĻŽ āĻšāϞ-

       āĻ•. āϝāĻ•ā§ƒāϤ

       āĻ–. āĻ…āĻ—ā§āĻ¨ā§āϝāĻžāĻļ⧟

       āĻ—. āĻĒā§āϞ⧀āĻšāĻž

       āϘ. āφāχāϞ⧇āϟāϏ āĻ…āĻŦ āĻ˛ā§āϝāĻžāĻ™ā§āĻ—āĻžāϰāĻšā§āϝāĻžāĻ¨ā§āϏ

 

āωāĻ¤ā§āϤāϰ : āϘ. āφāχāϞ⧇āϟāϏ āĻ…āĻŦ āĻ˛ā§āϝāĻžāĻ™ā§āĻ—āĻžāϰāĻšā§āϝāĻžāĻ¨ā§āϏ

 

⧧⧍. āϕ⧋āύāϟāĻŋ āĻ•ā§āϞ⧁ āĻŽā§‹āĻ˛ā§āĻĄ?

       āĻ•. Penicillium

       āĻ–. Saprolegnia

       āĻ—. Agaricus

       āϘ. Helminthosporium

 

āωāĻ¤ā§āϤāϰ : āĻ•. Penicillium

 

ā§§ā§Š. āĻ…āĻ¸ā§āĻĨāĻŋ āϝ⧇ āφāĻŦāϰāĻŖ āĻĻā§āĻŦāĻžāϰāĻž āφāĻŦ⧃āϤ āĻĨāĻžāϕ⧇ āϤāĻžāϕ⧇ āĻŦāϞ⧇-

       āĻ•. āĻĒ⧇āϰāĻŋāĻŸā§‹āύāĻŋ⧟āĻžāĻŽ

       āĻ–. āĻĒ⧇āϰāĻŋāĻ…āĻ¸ā§āϟāĻŋ⧟āĻžāĻŽ

       āĻ—. āĻĒ⧇āϰāĻŋāĻ•āĻžāĻ°ā§āĻĄāĻŋ⧟āĻžāĻŽ

       āϘ. āĻĒ⧇āϰāĻŋāĻ•āĻ¨ā§āĻĄā§āϰāĻŋ⧟āĻžāĻŽ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āĻĒ⧇āϰāĻŋāĻ…āĻ¸ā§āϟāĻŋ⧟āĻžāĻŽ

 

ā§§ā§Ē. āĻāĻŋāύ⧁āϕ⧇āϰ āĻ–ā§‹āϞāϕ⧇āϰ āϏāĻŦāĻšā§‡ā§Ÿā§‡ āĻ­āĻŋāϤāϰ⧇āϰ āĻ¸ā§āϤāϰ⧇āϰ āύāĻžāĻŽ āĻšāϞ-

       āĻ•. āĻ¨ā§āϝāĻžāĻ•āĻžāϰ āĻ¸ā§āϤāϰ

       āĻ–. āĻĒā§āϰāĻŋāϜāĻŽā§āϝāĻžāϟāĻŋāĻ• āĻ¸ā§āϤāϰ

       āĻ—. āĻĒ⧇āϰāĻŋāĻ“āĻ¸ā§āĻŸā§āϰāĻžāĻ•āĻžāĻŽ

       āϘ. āĻŽā§āϝāĻžāĻ¨ā§āϟāϞ

 

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻ¨ā§āϝāĻžāĻ•āĻžāϰ āĻ¸ā§āϤāϰ

 

ā§§ā§Ģ. āύāĻŋāωāĻ•ā§āϞāĻŋ⧟āĻžāϏ⧇āϰ āĻĒā§āϰāĻĨāĻŽ āĻŦāĻ°ā§āĻŖāύāĻž āĻ•āϰ⧇āύ-

      āĻ•. āϰāĻŦāĻžāĻ°ā§āϟ āĻŦā§āϰāĻžāωāύ

       āĻ–. āĻ…ā§āϝāĻžāϰāĻŋāĻ¸ā§āϟāϟāϞ

       āĻ—. āϞ⧁āχ āĻĒāĻžāĻ¸ā§āϤ⧁āϰ

       āϘ. āφāϞ⧇āĻ•āϜāĻžāĻ¨ā§āĻĄāĻžāϰ āĻĢā§āϞ⧇āĻŽāĻŋāĻ‚

 

āωāĻ¤ā§āϤāϰ : āĻ•. āϰāĻŦāĻžāĻ°ā§āϟ āĻŦā§āϰāĻžāωāύ

 

ā§§ā§Ŧ. āĻĢ⧁āϏāĻĢ⧁āĻ¸ā§€ā§Ÿ āϧāĻŽāύ⧀ āĻŦāĻšāύ āĻ•āϰ⧇-

       āĻ•. oxygenetaed blood

       āĻ–. deoxygenated blood

       āĻ—. pure blood

       āϘ. venous blood

 

āωāĻ¤ā§āϤāϰ : āĻ–. deoxygenated blood

 

ā§§ā§­. āĻ•āĻ•āϞāĻŋ⧟āĻž āĻšāϞ-

       āĻ•. āĻĻ⧃āĻˇā§āϟāĻŋāϰ āϏāĻžāĻĨ⧇ āϏāĻ‚āĻļā§āϞāĻŋāĻˇā§āϟ āĻ…āĻ™ā§āĻ—

       āĻ–. āĻļā§āϰāĻŦāϪ⧇āϰ āϏāĻžāĻĨ⧇ āϏāĻ‚āĻļā§āϞāĻŋāĻˇā§āϟ āĻ…āĻ™ā§āĻ—

       āĻ—. āĻ˜ā§āϰāĻžāϪ⧇āϰ āϏāĻžāĻĨ⧇ āϏāĻ‚āĻļā§āϞāĻŋāĻˇā§āϟ āĻ…āĻ™ā§āĻ—

       āϘ. āĻ–āĻžāĻĻā§āϝ āϚāĻ°ā§āĻŦāϪ⧇āϰ āϏāĻžāĻĨ⧇ āϏāĻ‚āĻļā§āϞāĻŋāĻˇā§āϟ āĻ…āĻ™ā§āĻ—

 

āωāĻ¤ā§āϤāϰ : āĻ–. āĻļā§āϰāĻŦāϪ⧇āϰ āϏāĻžāĻĨ⧇ āϏāĻ‚āĻļā§āϞāĻŋāĻˇā§āϟ āĻ…āĻ™ā§āĻ—

 

ā§§ā§Ž. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āϕ⧇āĻŦāϞāĻŽāĻžāĻ¤ā§āϰ āĻŦāĻžāĻ‚āϞāĻžāĻĻ⧇āĻļ⧇ āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ?

       āĻ•. Knema bengalensis

       āĻ–. Tactaria chattagramica

       āĻ—. Artocarpus heterophyllus

       āϘ. Ficus benghalensis

 

āωāĻ¤ā§āϤāϰ : āĻ•. Knema bengalensis

 

⧧⧝. 'Diversity and classification of flowering plants' āĻŦāχāϟāĻŋāϰ āϞ⧇āĻ–āĻ• āϕ⧇?

 

       āĻ•. Carolus Linnaeus

       āĻ–. George Bentham

       āĻ—. Theophrastus

       āϘ. Armen Takhtajan

 

āωāĻ¤ā§āϤāϰ : āϘ. Armen Takhtajan

 

⧍ā§Ļ. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋāϤ⧇ āĻĒā§āϞāĻžāϏāĻŽāĻŋāĻĄ āύ⧇āχ?

       āĻ•. E. coli

       āĻ–. A. Tumefaciens

       āĻ—. Yeast

       āϘ. Chlorella

 

āωāĻ¤ā§āϤāϰ : āĻ—. Yeast

 

⧍⧧. āĻŽāĻžāύ⧁āώ⧇āϰ RBC āĻāϰ āĻ—ā§œ āĻ†ā§Ÿā§āĻˇā§āĻ•āĻžāϞ āĻšāĻšā§āϛ⧇-

       āĻ•. 96 days

       āĻ–. 120 days

       āĻ—. 28 days

       āϘ. 62 days

 

āωāĻ¤ā§āϤāϰ : āĻ–. 120 days

 

⧍⧍. āϕ⧋āύāϟāĻŋ āϏāĻžā§ŸāĻžāύ⧋āĻŦā§āϝāĻžāĻ•āĻŸā§‡āϰāĻŋ⧟āĻž āύ⧟?

       āĻ•. Nostoc

       āĻ–. Anabaena

       āĻ—. Aulosira

       āϘ. Clostridium

 

āωāĻ¤ā§āϤāϰ : āϘ. Clostridium

 

ā§¨ā§Š. āϏāĻĒā§āϤāĻŽ āĻ•āϰ⧋āϟāĻŋāĻ•āĻž āĻ¸ā§āύāĻžā§Ÿā§āϕ⧇ āĻŦāϞāĻž āĻšā§Ÿ-

       āĻ•. āϭ⧇āĻ—āĻžāϏ

       āĻ–. āĻŸā§āϰāĻ•ā§āϞāĻŋ⧟āĻžāϰ

       āĻ—. āĻĢā§āϝāĻžāϏāĻŋ⧟āĻžāϞ

       āϘ. āĻ…āĻĒāϟāĻŋāĻ•

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻĢā§āϝāĻžāϏāĻŋ⧟āĻžāϞ

 

⧍ā§Ē. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋāϕ⧇ āĻšāĻŋāϞ āĻŦāĻŋāĻ•ā§āϰāĻŋ⧟āĻž āĻŦāϞāĻž āĻšā§Ÿ?

       āĻ•. CO2 āĻĨ⧇āϕ⧇ O2 āύāĻŋāĻ°ā§āĻ—āĻŽāύ

       āĻ–. H2O āĻĨ⧇āϕ⧇ O2 āύāĻŋāĻ°ā§āĻ—āĻŽāύ

       āĻ—. S2O āĻĨ⧇āϕ⧇ O2 āύāĻŋāĻ°ā§āĻ—āĻŽāύ

       āϘ. NO2 āĻĨ⧇āϕ⧇ O2 āύāĻŋāĻ°ā§āĻ—āĻŽāύ

 

āωāĻ¤ā§āϤāϰ : āĻ–. H2O āĻĨ⧇āϕ⧇ O2 āύāĻŋāĻ°ā§āĻ—āĻŽāύ

 

⧍ā§Ģ. āχāϞ⧇āϟāĻžāϰ⧇āϰ āĻ•āĻžāϜ āϕ⧀?

       āĻ•. āĻ–āĻžāĻĻā§āϝ āϤ⧈āϰ⧀ āĻ•āϰāĻž

       āĻ–. āĻ–āĻžāĻĻā§āϝ āϏāĻžā§āϚ⧟ āĻ•āϰāĻž

       āĻ—. āĻ¸ā§āĻĒā§‹āϰ āύāĻŋāĻ°ā§āĻ—āĻŽāύ āϏāĻžāĻšāĻžāĻ¯ā§āϝ āĻ•āϰāĻž

       āϘ. āĻŦāĻ‚āĻļ āĻŦāĻŋāĻ¸ā§āϤāĻžāϰ⧇ āĻ…āĻ‚āĻļāĻ—ā§āϰāĻšāĻŖ āĻ•āϰāĻž

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻ¸ā§āĻĒā§‹āϰ āύāĻŋāĻ°ā§āĻ—āĻŽāύ āϏāĻžāĻšāĻžāĻ¯ā§āϝ āĻ•āϰāĻž

 

⧍ā§Ŧ. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻĢāĻžāĻ°ā§āĻŖ āĻĒāĻžāϤāĻžāϰ āύāĻžāĻŽ?

       āĻ•. Stomium

       āĻ–. Fronds

       āĻ—. Prothallus

       āϘ. Scale leave

 

āωāĻ¤ā§āϤāϰ : āĻ–. Fronds

 

⧍⧭. āĻŽā§‡āϏ⧋āĻ—ā§āϞāĻŋ⧟āĻž āĻĒāĻžāĻ“ā§ŸāĻž āϝāĻžā§Ÿ āϕ⧋āύ āĻĒā§āϰāĻžāĻŖā§€āϤ⧇?

       āĻ•. āĻŽāĻžāύ⧁āώ

       āĻ–. āφāϰāĻļā§‹āϞāĻž

       āĻ—. āĻšāĻžāχāĻĄā§āϰāĻž

       āϘ. āĻŽāĻžāĻ›

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻšāĻžāχāĻĄā§āϰāĻž

 

ā§¨ā§Ž. T2 āĻŦā§āϝāĻžāĻ•āĻŸā§‡āϰāĻŋāĻ“āĻĢāĻžā§Ÿā§‡ āĻ•ā§ŸāϟāĻŋ āĻœā§€āύ āĻĨāĻžāϕ⧇?

       āĻ•. 150

       āĻ–. 145

       āĻ—. 155

       āϘ. 160

 

āωāĻ¤ā§āϤāϰ :

 

⧍⧝. āĻ“ā§ŸāĻžāϞ⧇āϏ⧇āϰ āϞāĻžāχāύ āĻšāϞ⧋ āĻāĻ•āϟāĻŋ āĻ•āĻžāĻ˛ā§āĻĒāύāĻŋāĻ• āϏ⧀āĻŽāĻžāϰ⧇āĻ–āĻž ............ āĻāĻŦāĻ‚ .......... āĻ…āĻžā§āϚāϞ⧇āϰ āĻŽāĻ§ā§āϝ⧇ āĻ…āĻŦāĻ¸ā§āĻĨāĻŋāϤāĨ¤

      āĻ•. āĻ“āϰāĻŋā§Ÿā§‡āĻ¨ā§āϟ āĻ“ āĻŦāĻžāĻ‚āϞāĻžāĻĻ⧇āĻļ

       āĻ–. āχāωāϰ⧋āĻĒ āĻ“ āĻāĻļāĻŋ⧟āĻž

       āĻ—. āĻŦāĻžāĻ‚āϞāĻžāĻĻ⧇āĻļ āĻ“ āĻ­āĻžāϰāϤ

       āϘ. āϕ⧋āϰāĻŋ⧟āĻž āĻ“ āϜāĻžāĻĒāĻžāύ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻ“āϰāĻŋā§Ÿā§‡āĻ¨ā§āϟ āĻ“ āĻŦāĻžāĻ‚āϞāĻžāĻĻ⧇āĻļ

 

ā§Šā§Ļ. āϕ⧋āύāϟāĻŋ āϜāĻžāχāϞ⧇āĻŽ āϟāĻŋāĻ¸ā§āϝ⧁āϰ āĻ…āĻ‚āĻļ āύ⧟?

       āĻ•. āϏāĻŋāĻ­ āύāϞ

       āĻ–. āĻŸā§āĻ°ā§āϝāĻžāĻ•āĻŋāĻĄ

       āĻ—. āĻŸā§āĻ°ā§āϝāĻžāĻ•āĻŋ⧟āĻž

       āϘ. āωāĻĄ āĻĢāĻžāχāĻŦāĻžāϰ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āϏāĻŋāĻ­ āύāϞ

 

English

1. The synonym of 'embellish' is

(A) adorn

(B) shock

(C) impoverish

(D) destroy

 

Answer: (A) adorn

 

2. The verb of 'beauty' is

(A) beautician

(B) beautify

(C) beautiful

(D) beautification

 

Answer: (B) beautify

 

3. The idiom 'let things slide' means-

(A) ignore

(B) lose gradually

(C) reveal a secret

(D) set free

 

Answer: (A) ignore

 

4. Choose the correct spelling

(A) indegenus

(B) indiginous

(C) indigenous

(D) indigeneous

 

Answer: (C) indigenous

 

5. Why is the poet so sad to see the daffodils in 'The Daffodils'?

(A) The poet is sad because the flowers have not bloomed fully.

(B) The poet is sad because winter will come soon.

(C) The poet is sad because the flowers remind him of his own death.

(D) The poet is sad because flowers bring very hot weather.

 

Answer: (C) The poet is sad because the flowers remind him of his own death.

 

6. The correct translation of  ‘āϰāĻŦāĻŋāĻŦāĻžāϰ āĻšāχāϤ⧇ āĻŦ⧃āĻˇā§āϟāĻŋ āĻšāχāϤ⧇āĻ›āĻŋāĻ˛â€™-

(A) It was raining from Sunday.

(B) It has been raining from Sunday.

(C) It had been raining since Sunday.

(D) It rained since Sunday.

 

Answer: (C) It had been raining since Sunday.

 

7. The word 'constraint' means

(A) freedom

(B) limitaion

(C) plentiful

(D) endless

 

Answer: (B) limitaion

 

8. The expression 'to look after' means

(A) to take care

(B) to follow

(C) to imitate

(D) to gaze

 

Answer: (A) to take care

 

9. The antonym of ‘stubborn’ is

(A) agreeable                    (B) obstinate                     (C) difficult                        (D) irritable

 

Answer: (A) agreeable

 

10. What is the meaning of the expression “bottom of my heart”?

(A) close to my heart

(B) core of my heart

(C) lower part of my heart

(D) close to my liver

 

Answer: (B) core of my heart

 

11. The word ‘desperation’ is a/an

(A) adjective                     (B) verb                              (C) adverb                          (D) noun

 

Answer: noun

 

12. Why were the daffodils in Wordswoth’s ‘I Wandered Lonely as a Cloud’ dancing?

(A) The poet was day dreaming.

(B) The flowers had cheerful company.

(C) The sea waves beside them had gone wild.

(D) There was a strong wind.

 

Answer: (D) There was a strong wind.

 

Questions 13-22: Fill in the blank/blanks:

13. I’d like ------ information, please.

(A) an

(B) some

(C) few

(D) piece

 

Answer: (B) some

 

14. Yoga is ------ a good exercise for ------ breathing.

(A) taken, controlling

(B) treated, considerable

(C) not, slowly

(D) considered, controlling

 

Answer: (D) considered, controlling

 

15. A man ----- by a speeding bus while he was crossing the road.

(A) was run over

(B) was run down

(C) had been run

(D) has been run over

 

Answer: (A) was run over

 

16. He ----- prefers ----- speak very little.

(A) doesn’t, to

(B) himself, to

(C) himself, for

(D) does, for

 

Answer: (B) himself, to

 

17. It ----- heavily when he ----- up.

(A) has snowed, woke

(B) snows, wake

(C) was snowing, woke

(D) is snowing, wakes

 

Answer: (C) was snowing, woke

 

18. Misuse of ----- energy has ----- destruction.

(A) solar, shown

(B) renewable, increase

(C) nuclear, cause

(D) atomic, wreaked

 

Answer: (D) atomic, wreaked

 

19. There is ----- milk in the bottle.

(A) very few

(B) any

(C) very little

(D) many

 

Answer: (C) very little

 

20. I have ------ him to give ----- smoking.

(A) said, up

(B) talked, for

(C) told, up

(D) told, in

 

Answer: (C) told, up

 

21. The groom arrived at the community centre exactly ----- time.

(A) in

(B) for

(C) by

(D) on

 

Answer: (D) on

 

22. He hates ----- kept ------.

(A) to be, waiting

(B) being kept, wait

(C) to be, to wait

(D) to, waiting

 

Answer: (A) to be, waiting

 

Questions 23-25: Choose the correct sentence:

23.

(A) It is you who is to pay.

(B) It is you who are to leave.

(C) It is you who is late.

(D) It is you who has won the prize.

 

Answer: (B) It is you who are to leave.

 

24.

(A) He suspicioned that something wrong.

(B) He suspicious that something is wrong.

(C) He suspected that something was wrong.

(D) He suspect that something is wrong.

 

Answer: (C) He suspected that something was wrong.

25.

(A) The jury are arguing among themselves.

(B) The jury is arguing among themselves.

(C) The jury has argued among themselves.

(D) The jury has been arguing among themselves.

 

Answer: (A) The jury are arguing among themselves.

 

Read the passage and answer questions 26-30:

 

Cats are carnivorous mammals of the family Felidae. They cannot chew their food, and their teeth are adapted to stab, anchor, and cut flesh. All cats except the cheetah have strong, sharp, retractile claws. They are not adapted for long chases but prowl their prey on padded feet and try to overwhelm it in short dash or pounce. Big cats roam over a large area, usually alone but sometimes in family groups, for example, a pride of lions can contain as many as 37 individuals. Cats generally are nocturnal animals, the retina of their eyes made extra sensitive to light by a layer of guanine, which causes the eyes to shine in the dark. Tigers are largest of the cats. They are identified by their characteristic striped coat. They inhabit forests and grasslands in Asia where populations have suffered from haunting, deforestation, and demand for tiger parts in traditional medicine.

 

26. ‘Cats are nocturnal animals’ means

(A) cats have predatory eyesight

(B) cats are sensitive to smell

(C) cats haunt during daytime

(D) cats are active at night

 

Answer: (D) cats are active at night

 

27. A group of lions is called ‘pride of lions’, what is a group of dogs called?

(A) a pack of dogs

(B) a colony of dogs

(C) a flock of dogs

(D) a cluster of dogs

 

Answer: (A) a pack of dogs

 

28. ‘retractile claws’ in the passage means

(A) claws the are extra sharp

(B) claws that can change shape

(C) claws that can move inwards

(D) claws that are sensitive to feelings

 

Answer: (C) claws that can move inwards

 

29. Cats are carnivorous, but human beings are

(A) herbivorous

(B) omnivorous

(C) farinaceous

(D) cadaverous

 

Answer: (B) omnivorous

 

30. The synonym of ‘prowl’ is

(A) dive

(B) chase

(C) ornace

(D) stalk

 

Answer: (D) stalk

 

āĻŦāĻžāĻ‚āϞāĻž

 

ā§§. āĻĒāĻžāĻžā§āĻœā§‡āϰ⧀- āĻ•āĻŦāĻŋāϤāĻžā§Ÿ āϝāĻžāĻ¤ā§āϰ⧀āϰāĻž āϕ⧋āĻĨāĻžā§Ÿ āĻŦāϏ⧇ āĻ…āĻĒ⧇āĻ•ā§āώāĻž āĻ•āϰ⧇?

       āĻ•. āϤ⧀āϰ⧇

       āĻ–. āĻŽāĻžāĻ¸ā§āϤ⧁āϞ⧇

       āĻ—. āĻŦāĻ¨ā§āĻĻāϰ⧇

       āϘ. āĻĻāĻžāĻā§œā§‡

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻŦāĻ¨ā§āĻĻāϰ⧇

 

āĻŸā§€āĻ•āĻž: āϰāĻžāϤ āĻĒā§‹āĻšāĻžāĻŦāĻžāϰ āĻ•āϤ āĻĻ⧇āϰāĻŋ āĻĒāĻžāĻžā§āĻœā§‡āϰāĻŋ?/ āĻŦāĻ¨ā§āĻĻāϰ⧇ āĻŦāϏ⧇ āϝāĻžāĻ¤ā§āϰ⧀āϰāĻž āĻĻāĻŋāύ āĻ—ā§‹āύ⧇, (āϤ⧃āĻ¤ā§€ā§Ÿ āĻ¸ā§āϤāĻŦāĻ•)

 

⧍. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻļ⧁āĻĻā§āϧ āĻŦāĻžāύāĻžāύ?

       āĻ•. āĻ¸ā§āĻŦāĻžā§ŸāĻ¤ā§āĻ¤ā§āĻŦāĻļāĻžāϏāύ

       āĻ–. āϏāĻžā§ŸāĻ¤ā§āĻ¤ā§āĻŦāĻļāĻžāϏāύ

       āĻ—. āĻ¸ā§āĻŦāĻžā§ŸāĻ¤ā§āϤāĻļāĻžāϏāύ

       āϘ. āĻ¸ā§āĻŦāĻžā§ŸāĻ¤ā§āĻ¤ā§āĻŦāĻļāĻžāώāĻŖ

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻ¸ā§āĻŦāĻžā§ŸāĻ¤ā§āϤāĻļāĻžāϏāύ

 

ā§Š. āĻŦāĻŋāϞāĻžāϏ⧀- āĻ—āĻ˛ā§āĻĒ⧇ āωāύāĻŋāĻļ āĻļāϤāϕ⧇āϰ āϝ⧇ āϏāĻŽāĻžāϜ-āϏāĻ‚āĻ¸ā§āĻ•āĻžāϰ⧇āϰ āĻ•āĻĨāĻž āφāϛ⧇ āϤāĻžāρāϰ āύāĻžāĻŽ-

 

       āĻ•. āψāĻļā§āĻŦāϰāϚāĻ¨ā§āĻĻā§āϰ āĻŦāĻŋāĻĻā§āϝāĻžāϏāĻžāĻ—āϰ

       āĻ–. āϰāĻžāĻŽāĻŽā§‹āĻšāύ āϰāĻžā§Ÿ

       āĻ—. āĻ…āĻ•ā§āώ⧟āϕ⧁āĻŽāĻžāϰ āĻĻāĻ¤ā§āϤ

       āϘ. āĻ­ā§‚āĻĻ⧇āĻŦ āĻŽā§āĻ–ā§‹āĻĒāĻžāĻ§ā§āϝāĻžā§Ÿ

 

āωāĻ¤ā§āϤāϰ : āϘ. āĻ­ā§‚āĻĻ⧇āĻŦ āĻŽā§āĻ–ā§‹āĻĒāĻžāĻ§ā§āϝāĻžā§Ÿ

 

āĻŸā§€āĻ•āĻž: āφāĻŽāĻŋ āĻ­ā§‚āĻĻ⧇āĻŦāĻŦāĻžāĻŦ⧁āϰ āĻĒāĻžāϰāĻŋāĻŦāĻžāϰāĻŋāĻ• āĻĒā§āϰāĻŦāĻ¨ā§āϧ⧇āϰāĻ“ āĻĻā§‹āώ āĻĻāĻŋāĻŦ āύāĻž āĻāĻŦāĻ‚ āĻļāĻžāĻ¸ā§āĻ¤ā§āĻ°ā§€ā§Ÿ āĻŦāĻŋāϧāĻŋ-āĻŦā§āϝāĻŦāĻ¸ā§āĻĨāĻžāϰāĻ“ āύāĻŋāĻ¨ā§āĻĻāĻž āĻ•āϰāĻŋāĻŦ āύāĻžāĨ¤ (āĻļ⧇āώ āĻ…āύ⧁āĻšā§āϛ⧇āĻĻ/āĻĒā§āϝāĻžāϰāĻžāĻ—ā§āϰāĻžāĻĢ)

 

ā§Ē. āϏāĻ¨ā§āϧāĻŋāϘāϟāĻŋāϤ āϕ⧋āύ āĻļāĻŦā§āĻĻāϟāĻŋ āĻļ⧁āĻĻā§āϧ?

 

   āĻ•. āĻŦ⧃āĻšāĻĻāĻ‚āĻļ

       āĻ–. āϜāĻžāĻ¤ā§āϝāĻžāĻ­āĻŋāĻŽāĻžāύ

       āĻ—. āφāĻĻā§āϝāĻžāĻ¨ā§āϤ

       āϘ. āĻļāĻŋāϰ⧋āĻšā§āϛ⧇āĻĻ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻŦ⧃āĻšāĻĻāĻ‚āĻļ

 

ā§Ģ. āϤāĻžāĻŽā§āϰāĻļāĻžāϏāύ- āĻŽāĻžāύ⧇?

 

       āĻ•. āĻ¸ā§āĻŦ⧈āϰāĻļāĻžāϏāύ

       āĻ–. āĻ•āĻžāϞ⧋ āφāχāύ

       āĻ—. āϤāĻžāĻŽāĻžāϰ āĻĒāĻžāϤ⧇ āĻ–ā§‹āĻĻāĻžāχ āĻ•āϰāĻž āφāĻĻ⧇āĻļ

       āϘ. āĻāĻ•āϜāύ āϰāĻžāϜāĻžāϰ āύāĻžāĻŽ

 

āωāĻ¤ā§āϤāϰ : āĻ—. āϤāĻžāĻŽāĻžāϰ āĻĒāĻžāϤ⧇ āĻ–ā§‹āĻĻāĻžāχ āĻ•āϰāĻž āφāĻĻ⧇āĻļ

 

ā§Ŧ. āĻĻ⧇āĻšā§‡āϰ āĻŦ⧟āϏ āĻ…āύ⧁āϝāĻžā§Ÿā§€ āĻŽāύ⧇āϰ āĻŦ⧟āϏ āύāĻž āĻŦāĻžā§œāϞ⧇ āϤāĻžāϕ⧇ āĻŦāϞ⧇-

 

       āĻ•. āĻļāĻžāϰ⧀āϰāĻŋāĻ• āĻĒā§āϰāϤāĻŋāĻŦāĻ¨ā§āϧ⧀

       āĻ–. āĻŽāύ⧋āĻŦāĻŋāĻ•āĻžāϰāĻ—ā§āϰāĻ¸ā§āϤ

       āĻ—. āĻŽāĻžāύāϏāĻŋāĻ• āϰ⧋āĻ—ā§€

       āϘ. āĻŦ⧁āĻĻā§āϧāĻŋāĻĒā§āϰāϤāĻŋāĻŦāĻ¨ā§āϧ⧀

 

āωāĻ¤ā§āϤāϰ : āϘ. āĻŦ⧁āĻĻā§āϧāĻŋāĻĒā§āϰāϤāĻŋāĻŦāĻ¨ā§āϧ⧀

 

ā§­. āϕ⧋āύ āĻļāĻŦā§āĻĻāϟāĻŋ āĻļ⧁āĻĻā§āϧ?

       āĻ•. āϏāĻŽā§€āĻ•ā§āώāύ

       āĻ–. āϏāĻŽā§€āĻšā§€āύ

       āĻ—. āĻļāĻŋāϰāĻšā§āϛ⧇āĻĻ

       āϘ. āĻ­āĻžāĻˇā§āĻ•āϰ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āϏāĻŽā§€āĻšā§€āύ

 

ā§Ž. āĻŽāĻžāϤ⧃āĻšā§ƒāĻĻā§Ÿā§‡ āĻĒāĻ•ā§āώāĻĒāĻžāϤāĻŋāϤāĻž āύāĻžāχāĨ¤â€“ āĻŦāĻžāĻ•ā§āϝāϟāĻŋ āϕ⧋āύ āϰāϚāύāĻžāϰ āĻ…āĻ¨ā§āϤāĻ°ā§āĻ—āϤ?

 

       āĻ•. āĻ…āĻ°ā§āϧāĻžāĻ™ā§āĻ—ā§€

       āĻ–. āϝ⧌āĻŦāύ⧇āϰ āĻ—āĻžāύ

       āĻ—. āĻŦāĻŋāϞāĻžāϏ⧀

       āϘ. āĻ•āĻŽāϞāĻžāĻ•āĻžāĻ¨ā§āϤ⧇āϰ āĻĻāĻĒā§āϤāϰ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻ…āĻ°ā§āϧāĻžāĻ™ā§āĻ—ā§€

 

⧝. āĻŦāĻžāĻ‚āϞāĻž āĻŦā§āϝāĻžā§āϜāύāĻŦāĻ°ā§āϪ⧇āϰ āĻĒā§āϰāϤāĻŋāĻŦāĻ°ā§āϗ⧇āϰ āĻĒāĻžā§āϚāĻŽ āĻŦāĻ°ā§āϗ⧇āϰ āĻ§ā§āĻŦāύāĻŋāϟāĻŋ-

       āĻ•. āĻ˜ā§‹āώāĻ§ā§āĻŦāύāĻŋ

       āĻ–. āĻ…āĻ˜ā§‹āώāĻ§ā§āĻŦāύāĻŋ

       āĻ—. āĻŽāĻšāĻžāĻĒā§āϰāĻžāĻŖāĻ§ā§āĻŦāύāĻŋ

       āϘ. āύāĻžāϏāĻŋāĻ•ā§āϝāĻ§ā§āĻŦāύāĻŋ

 

āωāĻ¤ā§āϤāϰ : āϘ. āύāĻžāϏāĻŋāĻ•ā§āϝāĻ§ā§āĻŦāύāĻŋ

 

āϞāĻŋāĻ™ā§āĻ•: āĻ§ā§āĻŦāύāĻŋ āĻ“ āĻŦāĻ°ā§āĻŖ āĻĒā§āϰāĻ•āϰāĻŖ āĻ“ āωāĻšā§āϚāĻžāϰāĻŖāĻŦāĻŋāϧāĻŋ

 

ā§§ā§Ļ. āϝ⧇āĻŽāύ āĻ•āĻ°ā§āĻŽ āϤ⧇āĻŽāύ āĻĢāϞ- āĻ āĻŦāĻžāĻ•ā§āϝ⧇ āĻŦā§āϝāĻŦāĻšā§ƒāϤ āĻšā§Ÿā§‡āϛ⧇-

       āĻ•. āύāĻŋāĻ°ā§āϧāĻžāϰāĻ• āĻŦāĻŋāĻļ⧇āώāĻŖ

       āĻ–. āĻ•ā§āϰāĻŋ⧟āĻž āĻŦāĻŋāĻļ⧇āώāĻŖ

       āĻ—. āϏāĻžāĻĒ⧇āĻ•ā§āώ āϏāĻ°ā§āĻŦāύāĻžāĻŽ

       āϘ. āĻŦāĻŋāĻļ⧇āώāϪ⧇āϰ āĻŦāĻŋāĻļ⧇āώāĻŖ

 

āωāĻ¤ā§āϤāϰ : āĻ—. āϏāĻžāĻĒ⧇āĻ•ā§āώ āϏāĻ°ā§āĻŦāύāĻžāĻŽ

 

āϞāĻŋāĻ™ā§āĻ•: āĻĒāĻĻ āĻĒā§āϰāĻ•āϰāĻŖ

 

ā§§ā§§. āĻĒā§āϰāĻŽāĻĨ āϚ⧌āϧ⧁āϰ⧀ āϏāĻŽā§āĻĒāĻžāĻĻāĻŋāϤ āϏāĻŦ⧁āϜāĻĒāĻ¤ā§āϰ- āĻĒāĻ¤ā§āϰāĻŋāĻ•āĻž āĻ•āϤ āϏāĻžāϞ⧇ āĻĒā§āϰāĻ•āĻžāĻļāĻŋāϤ āĻšā§Ÿ?

       āĻ•. ā§§ā§¯ā§§ā§Š

       āĻ–. ⧧⧝⧧ā§Ē

       āĻ—. ⧧⧝⧧ā§Ģ

       āϘ. ⧧⧝⧧⧭

 

āωāĻ¤ā§āϤāϰ : āĻ–. ⧧⧝⧧ā§Ē

 

⧧⧍. āϊāĻ°ā§āĻŖāĻžāϜāĻžāϞ- āĻļāĻŦā§āĻĻ⧇āϰ āĻ…āĻ°ā§āĻĨ

 

       āĻ•. āĻĻā§‹āĻĒāĻžāĻŸā§āϟāĻž

       āĻ–. āϕ⧁āĻœā§āĻāϟāĻŋāĻ•āĻž

       āĻ—. āĻŽāĻžāĻ•ā§œāϏāĻžāϞ āϤ⧈āϰāĻŋ āϜāĻžāϞ

       āϘ. āĻŽāĻžāĻ›āϧāϰāĻžāϰ āύāĻŋāĻ•ā§āώ⧇āĻĒāϝ⧋āĻ—ā§āϝ āϜāĻžāϞ

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻŽāĻžāĻ•ā§œāϏāĻžāϞ āϤ⧈āϰāĻŋ āϜāĻžāϞ

 

ā§§ā§Š. āĻĒāĻžāĻ ā§āϝ āϕ⧋āύ āĻ•āĻŦāĻŋāϤāĻžā§Ÿ āĻŦāĻžāϤāĻžāĻŦāĻŋ āύ⧇āĻŦ⧁-āϰ āωāĻ˛ā§āϞ⧇āĻ– āφāϛ⧇?

 

      āĻ•. āϤāĻžāĻšāĻžāϰ⧇āχ āĻĒā§œā§‡ āĻŽāύ⧇

       āĻ–. āĻŦāĻžāĻ‚āϞāĻžāĻĻ⧇āĻļ

       āĻ—. āĻāĻ•āϟāĻŋ āĻĢāĻŸā§‹āĻ—ā§āϰāĻžāĻĢ

       āϘ. āĻ•āĻŦāϰ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āϤāĻžāĻšāĻžāϰ⧇āχ āĻĒā§œā§‡ āĻŽāύ⧇

āĻŸā§€āĻ•āĻž: āĻŦāĻžāϤāĻžāĻŦāĻŋ āύ⧇āĻŦ⧁āϰ āĻĢ⧁āϞ āĻĢ⧁āĻŸā§‡āϛ⧇ āĻ•āĻŋ? āĻĢ⧁āĻŸā§‡āϛ⧇ āĻ•āĻŋ āφāĻŽā§‡āϰ āĻŽā§āϕ⧁āϞ? (āĻĒā§āϰāĻĨāĻŽ āĻ¸ā§āϤāĻŦāĻ•)

 

ā§§ā§Ē. āĻāĻ•āĻžāϧāĻŋāĻ• āĻ¸ā§āĻŦāĻžāϧ⧀āύ āĻŦāĻžāĻ•ā§āϝāϕ⧇ āĻāĻ•āϟāĻŋ āĻŦāĻžāĻ•ā§āϝ⧇ āϞāĻŋāĻ–āϞ⧇ āϏ⧇āϗ⧁āϞ⧋āϰ āĻŽāĻžāĻāĻ–āĻžāύ⧇ āϕ⧀ āϚāĻŋāĻšā§āύ āĻŦā§āϝāĻŦāĻšā§ƒāϤ āĻšā§Ÿ?

 

       āĻ•. āϕ⧋āϞāύ

       āĻ–. āĻĄā§āϝāĻžāĻļ

       āĻ—. āϏ⧇āĻŽāĻŋāϕ⧋āϞāύ

       āϘ. āĻ•āĻŽāĻž

 

āωāĻ¤ā§āϤāϰ : āϘ. āĻ•āĻŽāĻž

 

ā§§ā§Ģ. āφāĻŽāĻžāĻĻ⧇āϰ āĻĻāϞ⧇ āϕ⧇āĻš āĻĻāϞāĻĒāϤāĻŋ āύāĻžāχāĨ¤- āĻ•āĻžāĻšāĻžāĻĻ⧇āϰ āĻĻāϞ⧇?

 

       āĻ•. āĻ•āĻŦāĻŋāĻĻ⧇āϰ

       āĻ–. āϤāϰ⧁āĻŖāĻĻ⧇āϰ

       āĻ—. āϏāĻžāϧāĻ•āĻĻ⧇āϰ

       āϘ. āĻŦāĻ•ā§āϤāĻžāĻĻ⧇āϰ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āϤāϰ⧁āĻŖāĻĻ⧇āϰ

 

āĻŸā§€āĻ•āĻž: āφāĻŽāĻžāĻĻ⧇āϰ āĻĻāϞ⧇ āϕ⧇āĻš āĻĻāϞāĻĒāϤāĻŋ āύāĻžāχ, āφāϜ āφāĻŽāϰāĻž āĻļāϤ āĻĻāĻŋāĻ• āĻšāχāϤ⧇ āĻļāϤ āĻļāϤ āϤāϰ⧁āĻŖ āĻŽāĻŋāϞāĻŋ⧟āĻž āϤāĻžāϰ⧁āĻŖā§āϝ⧇āϰ āĻļāϤāĻĻāϞ āĻĢ⧁āϟāĻžāĻ‡ā§ŸāĻž āϤ⧁āϞāĻŋ⧟āĻžāĻ›āĻŋāĨ¤ (āϚāϤ⧁āĻ°ā§āĻĨ āĻ…āύ⧁āĻšā§āϛ⧇āĻĻ)

 

ā§§ā§Ŧ. The situation has come to a head- āĻāϰ āĻ…āĻ°ā§āĻĨ-

 

       āĻ•. āĻĒāϰāĻŋāĻ¸ā§āĻĨāĻŋāϤāĻŋāϰ āωāĻ¨ā§āύāϤāĻŋ āϘāĻŸā§‡āϛ⧇

       āĻ–. āĻĒāϰāĻŋāĻ¸ā§āĻĨāĻŋāϤāĻŋ āϏāĻŦāĻšā§‡ā§Ÿā§‡ āĻ­āĻžāϞ⧋ āĻ…āĻŦāĻ¸ā§āĻĨāĻžā§Ÿ āĻāϏ⧇ āĻĻāĻžāρ⧜āĻŋā§Ÿā§‡āϛ⧇

       āĻ—. āĻĒāϰāĻŋāĻ¸ā§āĻĨāĻŋāϤāĻŋāϰ āĻ…āĻŦāύāϤāĻŋ āϘāĻŸā§‡āϛ⧇

       āϘ. āĻĒāϰāĻŋāĻ¸ā§āĻĨāĻŋāϤāĻŋ āϚāϰāĻŽ āĻ…āĻŦāĻ¸ā§āĻĨāĻžā§Ÿ āĻĒ⧇⧗āρāϛ⧇āϛ⧇

 

āωāĻ¤ā§āϤāϰ : āϘ. āĻĒāϰāĻŋāĻ¸ā§āĻĨāĻŋāϤāĻŋ āϚāϰāĻŽ āĻ…āĻŦāĻ¸ā§āĻĨāĻžā§Ÿ āĻĒ⧇⧗āρāϛ⧇āϛ⧇

 

ā§§ā§­. āĻŽāĻŖā§āĻĄāĻ•- āĻļāĻŦā§āĻĻ⧇āϰ āĻ…āĻ°ā§āĻĨ-

       āĻ•. āĻŽā§āĻŖā§āĻĄāϧāĻžāϰ⧀

       āĻ–. āϭ⧇āĻ•

       āĻ—. āĻ•ā§ā§Ÿā§‹

       āϘ. āĻŽāĻŋāĻˇā§āϟāĻŋāϜāĻžāĻ¤ā§€ā§Ÿ āĻĻā§āϰāĻŦā§āϝ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āϭ⧇āĻ•

 

ā§§ā§Ž. āύāĻŋāĻšā§‡āϰ āϕ⧋āύāϟāĻŋ āĻļāĻžāĻŽāϏ⧁āϰ āϰāĻžāĻšāĻŽāĻžāύ⧇āϰ āĻ•āĻžāĻŦā§āϝāĻ—ā§āϰāĻ¨ā§āĻĨ āύ⧟?

       āĻ•. āĻĻ⧁āσāϏāĻŽā§Ÿā§‡āϰ āĻŽā§āĻ–ā§‹āĻŽā§āĻ–āĻŋ

       āĻ–. āωāĻĻā§āĻ­āϟ āωāĻŸā§‡āϰ āĻĒāĻŋāϠ⧇ āϚāϞ⧇āϛ⧇ āĻ¸ā§āĻŦāĻĻ⧇āĻļ

       āĻ—. āύāĻŋāϜ āĻŦāĻžāϏāĻ­ā§‚āĻŽā§‡

       āϘ. āĻāĻ•āĻ• āϏāĻ¨ā§āĻ§ā§āϝāĻžā§Ÿ āĻŦāϏāĻ¨ā§āϤ

 

āωāĻ¤ā§āϤāϰ : āϘ. āĻāĻ•āĻ• āϏāĻ¨ā§āĻ§ā§āϝāĻžā§Ÿ āĻŦāϏāĻ¨ā§āϤ

 

⧧⧝. āϕ⧋āύ āĻļāĻŦā§āĻĻāϟāĻŋ āωāĻĒāϏāĻ°ā§āĻ—āϝ⧋āϗ⧇ āĻ—āĻ āĻŋāϤ?

 

       āĻ•. āĻ…āĻŦāϰ⧇āĻŖā§āϝ

       āĻ–. āϤāϰ⧁āĻŖ

       āĻ—. āĻĒāϰ⧀āĻ•ā§āώāĻž

       āϘ. āĻ•āϞ⧁āώ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻ…āĻŦāϰ⧇āĻŖā§āϝ

 

āϞāĻŋāĻ™ā§āĻ•: āωāĻĒāϏāĻ°ā§āĻ—

 

⧍ā§Ļ. āϏāĻžāĻšāϚāĻ°ā§āϝ- āĻļāĻŦā§āĻĻāϟāĻŋ āĻ—āĻ āĻŋāϤ āĻšā§Ÿā§‡āϛ⧇

 

      āĻ•. āĻĒā§āϰāĻ¤ā§āϝ⧟āϝ⧋āϗ⧇

       āĻ–. āϧāĻžāϤ⧁āϝ⧋āϗ⧇

       āĻ—. āϏāĻ¨ā§āϧāĻŋāϝ⧋āϗ⧇

       āϘ. āϏāĻŽāĻžāϏāϝ⧋āϗ⧇

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻĒā§āϰāĻ¤ā§āϝ⧟āϝ⧋āϗ⧇

 

āϞāĻŋāĻ™ā§āĻ•: āĻĒā§āϰāĻ•ā§ƒāϤāĻŋ-āĻĒā§āϰāĻ¤ā§āϝ⧟

 

⧍⧧. āύāĻŋāϰāĻžāĻ•āĻžāϰ- āĻļāĻŦā§āĻĻ⧇āϰ āϏāĻ¨ā§āϧāĻŋāĻŦāĻŋāĻšā§āϛ⧇āĻĻ āϕ⧇āĻžāύāϟāĻŋ?

 

       āĻ•. āύāĻŋ+āφāĻ•āĻžāϰ

       āĻ–. āύāĻŋāσ+āφāĻ•āĻžāϰ

       āĻ—. āύāĻŋāϰ+āφāĻ•āĻžāϰ

       āϘ. āύāĻŋāϰāσ+āφāĻ•āĻžāϰ

 

āωāĻ¤ā§āϤāϰ : āĻ–. āύāĻŋāσ+āφāĻ•āĻžāϰ

 

⧍⧍. āĻ•āĻ°ā§āĻŦā§‚āϰ- āĻļāĻŦā§āĻĻ⧇āϰ āĻ…āĻ°ā§āĻĨ

 

       āĻ•. āϰāĻžāĻ•ā§āώāϏ

       āĻ–. āĻ—āĻ¨ā§āϧāĻĻā§āϰāĻŦā§āϝāĻŦāĻŋāĻļ⧇āώ

       āĻ—. āϰāĻžāϏāĻžā§ŸāύāĻŋāĻ• āĻĒāĻĻāĻžāĻ°ā§āĻĨ

       āϘ. āĻ•āϰāĻŖā§€ā§Ÿ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āϰāĻžāĻ•ā§āώāϏ

 

ā§¨ā§Š. āϤ⧁āώāĻžāϰāϧāĻŦāϞ- āϕ⧋āύ āϏāĻŽāĻžāϏ⧇āϰ āωāĻĻāĻžāĻšāϰāĻŖ?

 

       āĻ•. āϏāĻžāϧāĻžāϰāĻŖ āĻ•āĻ°ā§āĻŽāϧāĻžāϰ⧟

       āĻ–. āωāĻĒāĻŽāĻžāύ āĻ•āĻ°ā§āĻŽāϧāĻžāϰ⧟

       āĻ—. āωāĻĒāĻŽāĻŋāϤ āĻ•āĻ°ā§āĻŽāϧāĻžāϰ⧟

       āϘ. āĻŽāĻ§ā§āϝāĻĒāĻĻāϞ⧋āĻĒā§€ āĻ•āĻ°ā§āĻŽāϧāĻžāϰ⧟

 

āωāĻ¤ā§āϤāϰ : āĻ–. āωāĻĒāĻŽāĻžāύ āĻ•āĻ°ā§āĻŽāϧāĻžāϰ⧟

 

āϞāĻŋāĻ™ā§āĻ•: āϏāĻŽāĻžāϏ

 

⧍ā§Ē. āĻŖ-āĻ¤ā§āĻŦ āĻŦāĻŋāϧāĻŋ āĻ…āύ⧁āϏāĻžāϰ⧇ āϕ⧋āύ āϗ⧁āĻšā§āĻ› āĻ…āĻļ⧁āĻĻā§āϧ āĻŦāĻžāύāĻžāύ⧇āϰ āĻĻ⧃āĻˇā§āϟāĻžāĻ¨ā§āϤ?

       āĻ•. āϧāϰāύ, āĻŦāϰāĻŖ

       āĻ–. āĻŦāĻ°ā§āύāύāĻž, āĻĒ⧁āϰ⧋āύ⧋

       āĻ—. āύ⧇āĻ¤ā§āϰāϕ⧋āύāĻž, āĻĒāϰāĻ—āύāĻž

       āϘ. āϰ⧂āĻĒāĻžā§ŸāĻŖ, āĻĒā§āϰāϪ⧟āύ

āĻŸā§€āĻ•āĻž: āĻĻā§āĻŦāĻŋāĻ¤ā§€ā§Ÿ āϗ⧁āĻšā§āϛ⧇āϰ āĻļ⧁āĻĻā§āϧāϰ⧂āĻĒ- āĻŦāĻ°ā§āĻŖāύāĻž, āĻĒ⧁āϰāύ⧋

 

āωāĻ¤ā§āϤāϰ : āĻ–. āĻŦāĻ°ā§āύāύāĻž, āĻĒ⧁āϰ⧋āύ⧋

 

āϞāĻŋāĻ™ā§āĻ•: āĻŖāĻ¤ā§āĻŦ āĻ“ āώāĻ¤ā§āĻŦ āĻŦāĻŋāϧāĻžāύ

 

⧍ā§Ģ. āϏāĻŽāĻ­āĻŋāĻŦā§āϝāĻžāĻšāĻžāϰ- āĻļāĻŦā§āĻĻāϟāĻŋāϤ⧇ āĻŽā§‹āϟ āĻ•ā§ŸāϟāĻŋ āωāĻĒāϏāĻ°ā§āĻ— āφāϛ⧇?

       āĻ•. ⧍

       āĻ–. ā§Š

       āĻ—. ā§§

       āϘ. ā§Ē

āĻŸā§€āĻ•āĻž: āϏāĻŽ+āĻ…āĻ­āĻŋ+āĻŦāĻŋ+āĻšāĻžāϰ (āϏāĻŽ, āĻ…āĻ­āĻŋ, āĻŦāĻŋ- ā§ŠāϟāĻŋ āωāĻĒāϏāĻ°ā§āĻ—)

 

āωāĻ¤ā§āϤāϰ : āĻ–. ā§Š

 

āϞāĻŋāĻ™ā§āĻ•: āωāĻĒāϏāĻ°ā§āĻ—

 

⧍ā§Ŧ. āĻ•āĻžāĻŦā§āϝ āϜāĻ—āϤ⧇ āϝāĻžāϰ āύāĻžāĻŽ āφāύāĻ¨ā§āĻĻ āϤāĻžāϰ⧇āχ āύāĻžāĻŽ āĻŦ⧇āĻĻāύāĻžāĨ¤â€“ āĻŦāĻžāĻ•ā§āϝāϟāĻŋ āφāϛ⧇ āϝ⧇ āϰāϚāύāĻžā§Ÿ-

       āĻ•. āĻŦāĻŋāϞāĻžāϏ⧀

       āĻ–. āĻšā§ˆāĻŽāĻ¨ā§āϤ⧀

       āĻ—. āϏāĻžāĻšāĻŋāĻ¤ā§āϝ⧇ āϖ⧇āϞāĻž

       āϘ. āϝ⧌āĻŦāύ⧇āϰ āĻ—āĻžāύ

 

āωāĻ¤ā§āϤāϰ :  āĻ—. āϏāĻžāĻšāĻŋāĻ¤ā§āϝ⧇ āϖ⧇āϞāĻž

 

⧍⧭. āĻ•āĻžāϰ āĻŽā§‡āĻ›ā§‹ā§ŸāĻžāĻ• āĻ•āϰāĻž āĻĻāĻžāρāϤ āĻŦāĻžāϞāĻŦ-āĻāϰ āφāĻ˛ā§‹ā§Ÿ āĻāĻ•āĻāĻ• āĻ•āϰ⧇?

 

       āĻ•. āĻŽā§‹āĻĻāĻžāĻŦā§āĻŦ⧇āϰ⧇āϰ

       āĻ–. āĻŽāĻ•āϏ⧁āĻĻ⧇āϰ

       āĻ—. āχāωāύ⧁āϏ⧇āϰ

       āϘ. āϰāĻžāĻšāĻžāϤ⧇āϰ

 

āωāĻ¤ā§āϤāϰ : āĻ•. āĻŽā§‹āĻĻāĻžāĻŦā§āĻŦ⧇āϰ⧇āϰ

 

āĻŸā§€āĻ•āĻž: -āĻ“āϰāĻžāχ āϤ⧋ āϏāĻŦāĻ•āĻŋāϛ⧁āϰ āĻŽā§‚āϞ⧇, āĻŽā§‹āĻĻāĻžāĻŦā§āĻŦ⧇āϰ āĻŦāϞ⧇āĨ¤ āωāϞāĻ™ā§āĻ— āĻŦāĻžāϞāĻŦ-āĻāϰ āφāĻ˛ā§‹ā§Ÿ āϤāĻžāϰ āϏāϝāĻ¤ā§āύ⧇ āĻŽā§‡āĻ›ā§‹ā§ŸāĻžāĻ• āĻ•āϰāĻž āĻĻāĻžāρāϤ āĻāĻ•āĻāĻ• āĻ•āϰ⧇āĨ¤ āϤāĻžāĻĻ⧇āϰ āύ⧀āϚāϤāĻž āĻšā§€āύāϤāĻž āĻ—ā§‹āρ⧜āĻžāĻŽāĻŋāϰ āϜāĻ¨ā§āϝ⧇āχ āϤ⧋ āĻĻ⧇āĻļāϟāĻž āĻ­āĻžāĻ— āĻšāϞ⧋āĨ¤

 

ā§¨ā§Ž. Lyric āĻļāĻŦā§āĻĻ⧇āϰ āĻŦāĻžāĻ‚āϞāĻž āĻĒāϰāĻŋāĻ­āĻžāώāĻž=

       āĻ•. āĻ—āĻžāύ

       āĻ–. āϏ⧁āϰ

       āĻ—. āĻ•āĻĨāĻž

       āϘ. āĻ—ā§€āϤāĻŋāĻ•āĻŦāĻŋāϤāĻž

 

āωāĻ¤ā§āϤāϰ : āϘ. āĻ—ā§€āϤāĻŋāĻ•āĻŦāĻŋāϤāĻž

 

⧍⧝. Vertical āĻļāĻŦā§āĻĻ⧇āϰ āĻĒāϰāĻŋāĻ­āĻžāώāĻž-

       āĻ•. āĻ…āύ⧁āĻ­ā§‚āĻŽāĻŋāĻ•

       āĻ–. āωāĻšā§āϚāϤāĻž

       āĻ—. āωāĻ˛ā§āϞāĻŽā§āĻŦ

       āϘ. āĻĻā§‡ā§ŸāĻžāϞ

 

āωāĻ¤ā§āϤāϰ : āĻ—. āωāĻ˛ā§āϞāĻŽā§āĻŦ

 

ā§Šā§Ļ.  āĻĒāĻŋāĻĒāĻžāϏāĻŋāϤ āĻļāĻŦā§āĻĻ⧇āϰ āĻŦāĻŋāĻļ⧇āĻˇā§āϝāϰ⧂āĻĒ-

       āĻ•. āĻĒāĻŋāĻĒāĻžāϏ⧀

       āĻ–. āĻĒāĻŋāĻĒāĻžāϏ⧁

       āĻ—. āĻĒāĻŋāĻĒāĻžāϏāĻž

       āϘ. āĻĒāĻŋ⧟āĻžāϏ⧀

 

āωāĻ¤ā§āϤāϰ : āĻ—. āĻĒāĻŋāĻĒāĻžāϏāĻž

 

5 & 11 no ans are incorrect.....5(āĻ—)   &  11 (āĻ•)

āĻļāύāĻŋ, 10/13/2012 - 22:47
madina manwara (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

vaia  sound  diffraction  show  kre  bt  plarisation  shoe  kre  na........source  shajahan tapan sir  er boi.......
 

āĻļāύāĻŋ, 10/13/2012 - 22:55
shawon (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

 
boltsman's constant er akok J/K (source phy 1,tofajjol sir)
r hidrogen poromanur shoktistor er ans to 10.2ev hoar kotha,as energy absorvation E=E2-E1?

āϰāĻŦāĻŋ, 10/14/2012 - 12:57
nabil (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

25.
(A) The jury are arguing among themselves.
(B) The jury is arguing among themselves.
(C) The jury has argued among themselves.
(D) The jury has been arguing among themselves.
 
Answer: (B) The jury is arguing among themselves.
 
Jury er por is boshe.MS word a dilei correct kore dibe.

āϰāĻŦāĻŋ, 10/14/2012 - 17:14
MOYEEN (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

Ans. A

āϏ⧋āĻŽ, 10/15/2012 - 20:51
Suravi (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

āĻ—āĻŖāĻŋāϤ āĻāϰ ⧍⧭ āύāĻ‚ āĻĒā§āϰāĻļā§āύ⧇āϰ āωāĻ¤ā§āϤāϰ āϭ⧁āϞ āĻ°ā§Ÿā§‡āϛ⧇ āĨ¤ āĻĒā§āϰāĻļā§āύ⧇ āĻāĻ•āϟāĻŋ āĻŽā§‹āϜāĻž āĻĒāϰāĻžāϰ āĻ•āĻĨāĻž āĻŦāϞāĻž āĻšā§Ÿā§‡āϛ⧇ āĨ¤ āĻ•āĻŋāĻ¨ā§āϤ⧁ āĻāχ āϏāĻŽāĻžāϧāĻžāύ⧇ āĻāĻ• āĻœā§‹āϰāĻž āĻŽā§‹āϜāĻž āĻšāĻŋāϏāĻžāĻŦ āĻ•āϰāĻž āĻšā§Ÿā§‡āϛ⧇ āĨ¤

āĻŦ⧁āϧ, 10/17/2012 - 08:12
Gourab (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

chemistry 17 number question er ans b hobe.
 

āĻŦ⧁āϧ, 10/17/2012 - 12:52
Arif (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

12.āĻ—) 10.2eV
13.6-3.4=10.2 eV

āĻŦ⧁āϧ, 10/17/2012 - 18:31
Ismail (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

20. 50/20=2.5

āĻŦ⧁āϧ, 10/17/2012 - 18:34
Ismail (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

bio 8 num ki thik ace????

āĻŦ⧁āϧ, 10/17/2012 - 22:02
sattyajit (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

13. I’d like ------ information, please.
(A) an
(B) some
(C) few
(D) piece
 
āĻāϟāĻž āϤ⧋ āĻŽāύ⧇ āĻšā§Ÿ Few āĻšā§ŸāĻžāϰ āĻ•āĻĨāĻžāĨ¤ āφāĻŽāĻžāϰ āϜāĻžāύāĻž āĻŽāϤ⧇ Information āĻāϰ āφāϗ⧇ Few āĻŦāϏ⧇āĨ¤

āϏ⧋āĻŽ, 10/22/2012 - 12:36
āφāĻŦā§āĻĻ⧁āϞ āĻšāĻžāϞāĻŋāĻŽ (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

ā§§ āύāĻ‚ āĻĒā§āϰāĻļā§āύ⧇āϰ āĻŦā§āϝāĻžāĻ–ā§āϝāĻžā§Ÿ āϭ⧁āϞ āĻ°ā§Ÿā§‡āϛ⧇ ⧎

āĻŽāĻ™ā§āĻ—āϞ, 10/23/2012 - 13:23
asif mahbub protik (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

MATH āφāϰ PHYSICS āϏāĻ āĻŋāĻ• āωāĻ¤ā§āϤāϰ⧇āϰ āĻĒāĻžāĻļāĻžāĻĒāĻžāĻļāĻŋ āϏāĻŽāĻžāϧāĻžāύ āĻĻāĻŋā§Ÿā§‡ āĻĻ⧇āĻ“ā§ŸāĻžāϤ⧇ āϧāĻ¨ā§āϝāĻŦāĻžāĻĻ!! 

āĻŦ⧁āϧ, 11/14/2012 - 02:17
āφāĻŦāĻŋāĻĻ āĻšāĻžāϏāĻžāύ (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

⧍⧝ āύāĻ‚ āĻĒā§āϰāĻļā§āύ⧇āϰ āωāĻ¤ā§āϤāϰ C J/k

āĻļāύāĻŋ, 11/02/2013 - 08:37
sujon (āϝāĻžāϚāĻžāχāĻ•ā§ƒāϤ āύ⧟)

āύāϤ⧁āύ āĻ•āĻŽā§‡āĻ¨ā§āϟ āϝ⧁āĻ•ā§āϤ āĻ•āϰ⧁āύ