1
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+1
-0

If there is a change of angular momentum from $$1 \mathrm{j}$$-$$\mathrm{s}$$ to $$4 \mathrm{j}$$-$$\mathrm{s}$$ in $$4 \mathrm{~s}$$, then the torque

A
$$\left(\frac{5}{4}\right) \mathrm{J}$$
B
$$\left(\frac{3}{4}\right) \mathrm{J}$$
C
$$1 \mathrm{~J}$$
D
$$\left(\frac{4}{3}\right) \mathrm{J}$$
2
MHT CET 2020 16th October Morning Shift
MCQ (Single Correct Answer)
+1
-0

A solid cylinder of radius $$r$$ and mass $$M$$ rolls down an inclined plane of height $$h$$. When it reaches the bottom of the plane, then its rotational kinetic energy is ($$g=$$ acceleration due to gravity)

A
$$\frac{M g h}{4}$$
B
$$\frac{M g h}{2}$$
C
$$M g h$$
D
$$\frac{M g h}{3}$$
3
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A rod $l \mathrm{~m}$ long is acted upon by a couple as shown in the figure. The moment of couple is $\tau \mathrm{~Nm}$. If the force at each end of the rod, then magnitude of each force is

$$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)$$

MHT CET 2019 3rd May Morning Shift Physics - Rotational Motion Question 54 English

A
$\frac{\tau}{l}$
B
$\frac{l}{2 \tau}$
C
$\frac{2 \tau}{l}$
D
$\frac{2}{\tau}$
4
MHT CET 2019 3rd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

A
$\sqrt{70} \mathrm{~m} / \mathrm{s}$
B
$\sqrt{\frac{140}{3}} \mathrm{~m} / \mathrm{s}$
C
$\sqrt{\frac{280}{3}} \mathrm{~m} / \mathrm{s}$
D
$\sqrt{100} \mathrm{~m} / \mathrm{s}$
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