1
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is -
A
$${F \over {2mR}}$$
B
$${2F \over {3mR}}$$
C
$${3F \over {2mR}}$$
D
$${F \over {3mR}}$$

Explanation



FR = $${3 \over 2}$$ MR2$$\alpha $$

$$\alpha $$ = $${{2F} \over {3MR}}$$
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Evening Slot

Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :

A
$${{17} \over {15}}$$ MR2
B
$${{137} \over {15}}$$ MR2
C
$${{209} \over {15}}$$ MR2
D
$${{152} \over {15}}$$ MR2

Explanation

For Ball

using parallel axis theorem.

Iball = $${2 \over 5}$$MR2 + M(2R)2

= $${{22} \over 5}$$ MR2

2 Balls   so  $${{44} \over 5}$$MR2

Irod = for rod $${{M{{(2R)}^2}} \over R}$$ = $${{M{R^2}} \over 3}$$

Isystem = IBall + Irod

= $${{44} \over 5}M{R^2} + {{M{R^2}} \over 3}$$

= $${{137} \over {15}}$$ MR2
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o around the centre of the circle?
A
zero
B
10 m/s
C
$$10\sqrt 2 m/s$$
D
$$10\sqrt 3 m/s$$

Explanation


$$\left| {\Delta \overrightarrow v } \right| = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos \left( {\pi - \theta } \right)} $$

$$ = 2v\sin {\theta \over 2}$$        since $$\left[ {\left| {\overline v {}_1} \right| = \left| {{{\overline v }_2}} \right|} \right]$$

$$ = \left( {2 \times 10} \right) \times \sin \left( {{{30}^o}} \right)$$

= 10 m/s
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 11th January Morning Slot

A slab is subjected to two forces $$\overrightarrow {{F_1}} $$ and $$\overrightarrow {{F_2}} $$ of same magnitude F as shown in the figure. Force $$\overrightarrow {{F_2}} $$ is in XY-plane while force $$\overrightarrow {{F_1}} $$ acts along z = axis at the point $$\left( {2\overrightarrow i + 3\overrightarrow j } \right).$$. The moment of these forces about point O will be :

A
$$\left( {3\widehat i - 2\widehat j - 3\widehat k} \right)F$$
B
$$\left( {3\widehat i + 2\widehat j - 3\widehat k} \right)F$$
C
$$\left( {3\widehat i + 2\widehat j + 3\widehat k} \right)F$$
D
$$\left( {3\widehat i - 2\widehat j + 3\widehat k} \right)F$$

Explanation

Torque for F1 force

$${\overrightarrow F _1}$$ = $${F \over 2}\left( { - \widehat i} \right) + {{F\sqrt 3 } \over 2}\left( { - \widehat j} \right)$$

$${\overrightarrow r _1} = 0\widehat i + 6\widehat j$$

$${\overrightarrow \tau _{_{{F_1}}}} = \overrightarrow {{r_1}} \times \overrightarrow {{F_1}} = 3F\widehat k$$

Torque for F2 force

$$\overrightarrow {{F_2}} = F\widehat k$$

$$\overrightarrow {{r_2}} = 2\widehat i + 3\widehat j$$

$${\overrightarrow \tau _{_{{F_2}}}} = \overrightarrow {{r_2}} \times \overrightarrow {{F_2}} = 3F\widehat i + 2F\left( { - \widehat j} \right)$$

$${\overrightarrow \tau _{net}} = {\overrightarrow \tau _{{F_1}}} + {\overrightarrow \tau _{{f_2}}}$$

$$ = 3F\widehat i + 2F\left( { - \widehat j} \right) + 3F\left( {\widehat k} \right)$$

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