NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

A body of mass m is thrown vertically upward with speed $$\sqrt3$$ ve, where ve is the escape velocity of a body from earth surface. The final velocity of the body is

A
0
B
2 ve
C
$$\sqrt3$$ ve
D
$$\sqrt2$$ ve

Explanation

Given, initial speed, $$v = \sqrt 3 \,{v_e}$$

We know that, $${v_e} = \sqrt {{{2GM} \over R}} $$

At the surface of earth,

Total energy $$ = PE + KE$$

$$ = - {{GMm} \over R} + {1 \over 2}m{v^2}$$

$$ = - {{GMm} \over R} + {1 \over 2}m{\left( {\sqrt 3 \sqrt {{{2GM} \over R}} } \right)^2}$$

$$ = - {{GMm} \over R} + {1 \over 2}3m \times \,{{2GM} \over R}$$

$$ = -{{GMm} \over R} + {{3GMm} \over R} = {{2GMm} \over R}$$

At infinity, total energy $$ = PE + KE$$

$$ = 0 + {1 \over 2}\,m{v^2}$$

From the principle of conservation of energy,

$${{2GMm} \over R} = {1 \over 2}m{v^2}$$

$$ \Rightarrow {{4GM} \over R} = {v^2}$$

$$ \Rightarrow v = \sqrt {{{4GM} \over R}} $$

$$ = \sqrt {2 \times {{2GM} \over R}} = \sqrt 2 \,{v_e}$$

m ভরের একটি বস্তুকে খাড়া উপরের দিকে $$\sqrt 3 {v_e}$$ বেগে ছোঁড়া হল। যেখানে $${v_e}$$ হল পৃথিবীপৃষ্ঠে বস্তুটির মুক্তিবেগ। বস্তুটির অন্তিম বেগ হবে

A
0
B
2 ve
C
$$\sqrt3$$ ve
D
$$\sqrt2$$ ve
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
An ideal gas of molar mass M is contained in a very tall vertical cylindrical column in the uniform gravitational field. Assuming the gas temperature to be T, the height at which the centre of gravity of the gas is located is (R$$\to$$universal gas constant)
A
$${{RT} \over g}$$
B
$${{RT} \over Mg}$$
C
MgR
D
RTg

Explanation

As the gravitational field is uniform, therefore centre of gravity and the centre of mass are at same location.

$$\therefore$$ The location of centre of mass is

$$h = {{\int_0^\infty {hdm} } \over {\int_0^\infty {dm} }}$$

$$h = {{\int_0^\infty {h\rho dh} } \over {\int_0^\infty {\rho dh} }}$$ ..... (i)

But from barometric formula and gas law,

$$\rho = {\rho _0}{e^{ - Mgh/RT}}$$

$$\therefore$$ Putting the value of $$\rho$$ in Eq. (i), we have

$$h = {{\int_0^\infty {h{\rho _0}{e^{( - Mgh/RT)}}dh} } \over {\int_0^\infty {{\rho _0}{e^{ - (Mgh/RT)}}dh} }} = {{RT} \over {Mg}}$$
একটি সুষম অভিকর্ষ ক্ষেত্রে খুব লম্বা, খাড়া চোঙের মধ্যে M আণবিক ভরসম্পন্ন আদর্শ গ্যাস আবদ্ধ আছে। গ্যাসের তাপমাত্রা T ধরে নিলে, সমগ্র গ্যাসের ভারকেন্দ্র যে উচ্চতায় থাকবে তার মান (R: সার্বজনীন গ্যাস ধ্রুবক)
A
$${{RT} \over g}$$
B
$${{RT} \over Mg}$$
C
MgR
D
RTg

Explanation

মহাকর্ষীয় ক্ষেত্রটি অভিন্ন, তাই মহাকর্ষীয় কেন্দ্র এবং ভরের কেন্দ্র একই অবস্থানে রয়েছে।

$$\therefore$$ ভরের কেন্দ্রের অবস্থান হল

$$h = {{\int_0^\infty {hdm} } \over {\int_0^\infty {dm} }}$$

$$h = {{\int_0^\infty {h\rho dh} } \over {\int_0^\infty {\rho dh} }}$$ ..... (i)

কিন্তু ব্যারোমেট্রিক সূত্র এবং গ্যাস আইন থেকে,

$$\rho = {\rho _0}{e^{ - Mgh/RT}}$$

$$\therefore$$ সমীকরণ (i) এ p এর মান বসিয়ে পাই

$$h = {{\int_0^\infty {h{\rho _0}{e^{( - Mgh/RT)}}dh} } \over {\int_0^\infty {{\rho _0}{e^{ - (Mgh/RT)}}dh} }} = {{RT} \over {Mg}}$$
3

WB JEE 2019

MCQ (Single Correct Answer)
English
Bengali
Assume that the earth moves around the sun in a circular orbit of radius R and there exists a planet which also move around the sun in a circular orbit with an angular speed twice as large as that of the earth. The earth of the orbit of the planet is
A
$${2^{ - 2/3}}$$ R
B
$${2^{ 2/3}}$$ R
C
$${2^{ - 1/3}}$$ R
D
$${R \over {\sqrt 2 }}$$

Explanation

According to the Kepler's third law

T2 $$ \propto $$ r3

where, T = time period of revolution

r = radius

Now, $${\left( {{{{T_E}} \over {{T_P}}}} \right)^2} = {\left( {{{{r_E}} \over {{r_P}}}} \right)^3} = {{{r_E}} \over {{r_P}}} = {\left( {{{{T_E}} \over {{T_P}}}} \right)^{2/3}}$$

$${{{r_E}} \over {{r_P}}} = {\left( {{{2\pi /{\omega _E}} \over {2\pi /{\omega _P}}}} \right)^{2/3}}$$ $$ \because $$ $$\left[ {T = {{2\pi } \over \omega }} \right]$$

$${{{r_E}} \over {{r_P}}} = {\left( {{{{\omega _P}} \over {{\omega _E}}}} \right)^{2/3}}$$

According to the question, $${\omega _P} = 2{\omega _E}$$ and $${r_E} = R$$

$$ \Rightarrow {R \over {{r_P}}} = {\left( {{{2{\omega _E}} \over {{\omega _E}}}} \right)^{2/3}}$$

$${R \over {{r_P}}} = {(2)^{2/3}}$$

$${r_P} = {R \over {{{(2)}^{2/3}}}} = R{(2)^{ - 2/3}}$$

ধরা যাক, পৃথিবী সূর্যকে ঘিরে R ব্যাসার্ধের একটি বৃত্তাকার পথে প্রদক্ষিণ করে এবং অন্য আর একটি গ্রহ আছে যা পৃথিবীর কৌণিক বেগের দ্বিগুণ কৌণিক বেগে সূর্যকে বৃত্তাকার পথে প্রদক্ষিণ করে। গ্রহটির ব্যাসার্ধ হল -

A
2$$-$$2/3 R
B
22/3 R
C
2$$-$$1/3 R
D
$${R \over {\sqrt 2 }}$$

Explanation

যেহেতু,

$${T^2} \propto {r^3} \Rightarrow r \propto {T^{2/3}} \Rightarrow r \propto {1 \over {{\omega ^{2/3}}}}$$

অতএব,

$${{{r_p}} \over {{r_e}}} = {\left( {{{{\omega _e}} \over {{\omega _p}}}} \right)^{2/3}} \Rightarrow {r_p} = {r_e}{\left( {{{{\omega _e}} \over {{\omega _p}}}} \right)^{2/3}} = R\,.\,{\left( {{\omega \over {2\omega }}} \right)^{2/3}} = R{2^{ - 2/3}}$$

4

WB JEE 2018

MCQ (Single Correct Answer)
English
Bengali
The ratio of accelerations due to gravity g1 : g2 on the surfaces of two planets is 5 : 2 and the ratio of their respective average densities $${{\rho _1}}$$ : $${{\rho _2}}$$ is 2 : 1. What is the ratio of respective escape velocities v1 : v2 from the surface of the planets?
A
5 : 2
B
$${\sqrt 5 }$$ : $${\sqrt 2 }$$
C
5 : 2$${\sqrt 2 }$$
D
25 : 4

Explanation

$$ \because $$ Escape velocity from a planet,

$${v_e} = \sqrt {{{2GM} \over R}} = \sqrt {{{2GM} \over {{R^2}}}} R$$

$$ = \sqrt {2gR} $$ ...(i)

According due to gravity

$$g = {{GM} \over {{R^2}}} = {{G{4 \over 3}\pi {R^3}.\rho } \over {{R^2}}} = {4 \over 3}G\pi R\rho $$

$$ \therefore $$ Radius $$R = {{3g} \over {4\pi G\rho }}$$ .... (ii)

From Eqs. (i) and (ii), we get

$${v_e} = \sqrt {2g.{{3g} \over {4\pi G\rho }}} = \sqrt {{3 \over 2}{{{g^2}} \over {\pi G\rho }}} $$

Thus, $${v_e} \propto {g \over {\sqrt \rho }}$$

$$ \therefore $$ $${{{v_{{e_1}}}} \over {{v_{{e_1}}}}} = {{{g_1}} \over {\sqrt {{\rho _1}} }} \times {{\sqrt {{\rho _2}} } \over {{g_2}}} = {5 \over 2} \times {1 \over {\sqrt 2 }}$$

(given, $${{{g_1}} \over {{g_2}}} = {5 \over 2}$$ and $${{{\rho _1}} \over {{\rho _2}}} = 2:1$$)

$$ = {5 \over {2\sqrt 2 }}$$

দুটি গ্রহের পৃষ্ঠতলে অভিকর্ষজ ত্বরণের অনুপাত $${g_1}:{g_2} = 5:2$$ এবং তাদের গড় ঘনত্বের অনুপাত $${\rho _1}:{\rho _2} = 2:1$$ । গ্রহ দুটির পৃষ্ঠ থেকে মুক্তি বেগের অনুপাত $${V_1}:{V_2}$$ কত হবে ?

A
$$5:2$$
B
$$\sqrt 5 :\sqrt 2 $$
C
$$5:2\sqrt 2 $$
D
$$25:4$$

Explanation

আমরা জানি, কোন গ্রহের পৃষ্ঠ থেকে মুক্তিবেগ $${V_e}$$ হলে,

$${V_e} = \sqrt {2gR} $$ এবং গ্রহের ঘনত্ব $$\rho $$ হলে, $$\rho = {{3g} \over {4\pi RG}} \Rightarrow R = {{3g} \over {4\pi \rho G}}$$

সুতরাং, $${V_e} = \sqrt {{{2g\,.\,3g} \over {4\pi \rho G}}} \Rightarrow {V_e} \propto \sqrt {{{{g^2}} \over \rho }} $$

$$\therefore$$ নির্ণেয় $${{{V_1}} \over {{V_2}}} = \sqrt {{{\left( {{{{g_1}} \over {{g_2}}}} \right)}^2}} \,.\,\sqrt {{{{\rho _2}} \over {{\rho _1}}}} = \sqrt {{{25} \over 4}} \,.\,\sqrt {{1 \over 2}} = \sqrt {{{25} \over 8}} = {5 \over {2\sqrt 2 }}$$

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12