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### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
A bullet of 10 g, moving with velocity v, collides head-on with the stationary bob of a pendulum and recoils with velocity 100 m/s. The length of the pendulum is 0.5 m and mass of the bob is 1 kg. The minimum value of v = ____________ m/s so that the pendulum describes a circle. (Assume the string to be inextensible and g = 10 m/s2)

## Explanation

$$V' = \sqrt {5gR} = \sqrt {5 \times 10 \times 0.5}$$

V' = 5 m/s

m1 V = m2 $$\times$$ 5 $$-$$ m1 $$\times$$ 100

$${{10} \over {1000}} \times V = 5 - {{10} \over {1000}} \times 100$$

V = 400 m/s
2

### JEE Main 2021 (Online) 26th August Morning Shift

Numerical
Consider a badminton racket with length scales as shown in the figure.

If the mass of the linear and circular portions of the badminton racket are same (M) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $${r \over 2}$$ distance from the end A of the handle will be ................ Mr2.

## Explanation

$$I = \left[ {{I_1} + M{{\left( {{5 \over 2}r} \right)}^2}} \right] + \left[ {{I_2} + M{{\left( {{{13r} \over 2}} \right)}^2}} \right]$$

$$= \left[ {{{M(36{r^2})} \over {12}} + {{M(25{r^2})} \over 4}} \right] + \left[ {{{M{r^2}} \over 2} + {{169M{r^2}} \over 4}} \right]$$

= 52 Mr2

Ans. 52.00
3

### JEE Main 2021 (Online) 25th July Morning Shift

Numerical
A body of mass 2 kg moving with a speed of 4 m/s. makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial peed. The speed of the two body centre of mass is $${x \over {10}}$$ m/s. Then the value of x is ___________.

## Explanation

pi = pf

2 $$\times$$ 4 = 2 $$\times$$ 1 + m2 $$\times$$ v2

m2v2 = 6 ..... (i)

by coefficient of restitution

$$1 = {{{v_2} - 1} \over 4} \Rightarrow {v_2} = 5$$ m/s

by (i)

m2 $$\times$$ 5 = 6

m2 = 1.2 kg

$${v_{cm}} = {{{m_1}{v_1} + {m_2}{v_2}} \over {{m_1} + {m_2}}}$$

$${v_{cm}} = {{2 \times 1 + 1.2 \times 5} \over {2 + 1.2}} = {8 \over {3.2}} = {{25} \over {10}}$$

x = 25
4

### JEE Main 2021 (Online) 22th July Evening Shift

Numerical
The position of the centre of mass of a uniform semi-circular wire of radius 'R' placed in x-y plane with its centre at the origin and the line joining its ends as x-axis is given by $$\left( {0,{{xR} \over \pi }} \right)$$. Then, the value of | x | is ______________.

## Explanation

Centre of mass of half ring is located at a distance $${{2R} \over \pi }$$ from centre of the ring on its axis of symmetry so position of centre of mass in the given question will be $$\left( {0,{{xR} \over \pi }} \right)$$

$$\Rightarrow$$ | x | = 2

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