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1

### WB JEE 2022

MCQ (More than One Correct Answer)
English
Bengali

A particle is moving in x-y plane according to $$\overrightarrow r = b\cos \omega t\widehat i + b\sin \omega t\widehat j$$, where $$\omega$$ is constant. Which of the following statement(s) is/are true?

A
$${E \over \omega }$$ is a constant where E is the total energy of the particle.
B
The trajectory of the particle in x-y plane is a circle.
C
In ax - ay plane, trajectory of the particle is an ellipse (ax, ay denotes the components of acceleration)
D
$$\overrightarrow a = {\omega ^2}\overrightarrow v$$

## Explanation

Given, trajectory of the particle,

$$\overrightarrow r = b\cos \omega t\widehat i + b\sin \omega t\widehat j$$

It's component can be written as

$$x = b\cos \omega t$$ ..... (i)

$$y = b\sin \omega t$$ ...... (ii)

Both Eqs. (i) and (ii) together represent the circle in x-y plane. Thus, the trajectory of the particle in x-y plane is circle.

Now,

$${{d\overrightarrow r } \over {dt}} = \omega b( - \sin \omega t\widehat i + \cos \omega t\widehat j)$$

$${{{d^2}\overrightarrow r } \over {d{t^2}}} = - b{\omega ^2}(\cos \omega t\widehat i + \sin \omega t\widehat j)$$

$$= - {\omega ^2}(\overrightarrow r )$$

Therefore, $$\overrightarrow a = - {\omega ^2}\overrightarrow r$$

Hence, option (c) and (d) are incorrect.

Energy of the particle in circular motion is given as $${1 \over 2}m{\omega ^2}{A^2}$$ (where, A is the maximum displacement from the centre), which is constant.

Hence, $${E \over \omega }$$ is also constant.

So, option (a) and (b) are correct.

একটি কণা x-y তলে $$\overrightarrow r = b\cos \omega t\widehat i + b\sin \omega t\widehat j$$ সূত্র অনুযায়ী সঞ্চারণশীল, যেখানে $$\omega$$ একটি ধ্রুবক। সেক্ষেত্রে নীচের কোন উক্তিটি বা কোন কোন উক্তিগুলি সঠিক ?

A
$${E \over \omega }$$ ধ্রুবক যেখানে E হল কণাটির মোট শক্তি
B
x-y তলে কণাটির সঞ্চারপথ বৃত্তাকার
C
ax - ay তলে কণাটির সঞ্চারপথ উপবৃত্তাকার (ax ও ay হল ত্বরণের উপাংশদ্বয়)
D
$$\overrightarrow a = {\omega ^2}\overrightarrow v$$
2

### WB JEE 2021

MCQ (More than One Correct Answer)
English
Bengali
The potential energy of a particle of mass 0.02 kg moving along X-axis is given by V = Ax (x $$-$$ 4) J, where x is in metre and A is a constant. Which of the following statements is/are correct?
A
The particle is acted upon by a constant force.
B
The particle executes simple harmonic motion.
C
The speed of the particle is maximum at x = 2 m.
D
The period of oscillation of the particle is $${\pi \over 5}sec$$.

## Explanation

Mass, m = 0.02 kg

Potential energy,

$$U = Ax(x - 4)J$$

$$\Rightarrow U = A{x^2} - 4Ax$$

Force, $$F = - {{dV} \over {dx}} = - {d \over {dx}}(A{x^2} - 4Ax)$$

$$\Rightarrow F = - 2Ax + 4A$$ .... (i)

This force is dependent on x, hence particle is not acted upon by a constant force.

From Eq. (i), it is clear that f $$\propto$$ $$-$$ x

Hence, particle executes simple hormonic motion. Speed of particle is maximum at equilibrium position, i.e. when F = 0

i.e. $$-$$2Ax + 4A = 0

$$\Rightarrow$$ x = 2m

Period of oscillation,

$$T = 2\pi \sqrt {{m \over k}}$$ .... (i)

From Eq. (i), F = $$-$$2Ax + 4A

Since, value of k will be obtained in terms of A.

Therefore, the value of time period will be also obtained in terms of A, hence option (d) is not correct.
x-অক্ষ বরাবর গতিশীল 0.02 kg ভরের একটি কণার স্থিতিশক্তি V = Ax (x - 4)J, যেখানে A একটি ধ্রুবক এবং x মিটার এককে প্রকাশিত। নিচের কোন উক্তি/উক্তিগুলি সত্য?
A
কণাটির উপর প্রযুক্ত বলের মান ধ্রুবক।
B
কণাটি সরল দোলগতি সম্পন্ন করে।
C
x = 2 m বিন্দুতে কণাটির বেগ সর্বাধিক।
D
কণাটির দোলগতির দোলনকাল $${\pi \over 5}sec$$.

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