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1

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

Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, $$\omega$$1 and $$\omega$$2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :
A
$${{{I_1}{I_2}} \over {({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$$
B
$${{{{({I_1} - {I_2})}^2}{\omega _1}{\omega _2}} \over {2({I_1} + {I_2})}}$$
C
$${{{I_1}{I_2}} \over {2({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$$
D
$${{{{({\omega _1} - {\omega _2})}^2}} \over {2({I_1} + {I_2})}}$$

## Explanation

From conservation of angular momentum we get

$${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega$$

$$\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$$

$${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$$

$${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$$

$${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$$

Solving above we get

$${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$$
2

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

A huge circular arc of length 4.4 ly subtends an angle '4s' at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second?

Given : 1 ly = 9.46 $$\times$$ 1015 m

1 AU = 1.5 $$\times$$ 1011 m
A
4.1 $$\times$$ 108 s
B
4.5 $$\times$$ 1010 s
C
3.5 $$\times$$ 106 s
D
7.2 $$\times$$ 108 s

## Explanation

$$R = {l \over \theta }$$

Time $$= {{4 \times 2\pi R} \over v} = {{4 \times 2\pi } \over v}\left( {{l \over \theta }} \right)$$

put l = 4.4 $$\times$$ 9.46 $$\times$$ 1015

v = 8 $$\times$$ 1.5 $$\times$$ 1011

$$\theta = {4 \over {3600}} \times {\pi \over {180}}$$ rad.

we get time = 4.5 $$\times$$ 1010 sec.
3

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

The solid cylinder of length 80 cm and mass M has a radius of 20 cm. Calculate the density of the material used if the moment of inertia of the cylinder about an axis CD parallel to AB as shown in figure is 2.7 kg m2. A
14.9 kg/m3
B
7.5 $$\times$$ 101 kg/m3
C
7.5 $$\times$$ 102 kg/m3
D
1.49 $$\times$$ 102 kg/m3

## Explanation

Parallel axis theorem

I = ICM + Md2

$$I = {{M{r^2}} \over 2} + M{\left( {{L \over 2}} \right)^2}$$

$$2.7 = M{{{{(0.2)}^2}} \over 2} + M{\left( {{{0.8} \over 2}} \right)^2}$$

$$2.7 = M\left[ {{2 \over {100}} + {{16} \over {100}}} \right]$$

M = 15 kg

$$\Rightarrow \rho = {M \over {\pi {r^2}L}} = {{15} \over {\pi {{(0.2)}^2} \times 0.8}}$$

= 0.1492 $$\times$$ 103
4

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

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Moment of inertia of a circular disc of mass 'M' and radius 'R' about X, Y axes (passing through its plane) and Z-axis which is perpendicular to its plane were found to be Ix, Iy and Iz respectively. The respectively radii of gyration about all the three axes will be the same.

Reason R : A rigid body making rotational motion has fixed mass and shape. In the light of the above statements, choose the most appropriate answer from the options given below :
A
Both A and R are correct but R is NOT the correct explanation of A.
B
A is not correct but R is correct.
C
A is correct but R is not correct.
D
Both A and R are correct and R is the correct explanation of A.

## Explanation

Iz = Ix + Iy (using perpendicular axis theorem) & I = mk2 (K : radius of gyration)

so, mKz2 = mKx2 + mKy2

Kz2 = Kx2 + Ky2

so radius of gyration about axes x, y & z won't be same hence assertion A is not correct reason R is correct statement (property of a rigid body)

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