MHT CET 2020 16th October Morning Shift | Rotational Motion Question 60 | Physics

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)

$$\frac{M g h}{4}$$

$$\frac{M g h}{2}$$

$$M g h$$

$$\frac{M g h}{3}$$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-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 3 English

$\frac{\tau}{l}$

$\frac{l}{2 \tau}$

$\frac{2 \tau}{l}$

$\frac{2}{\tau}$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-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 $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

$\sqrt{70} \mathrm{~m} / \mathrm{s}$

$\sqrt{\frac{140}{3}} \mathrm{~m} / \mathrm{s}$

$\sqrt{\frac{280}{3}} \mathrm{~m} / \mathrm{s}$

$\sqrt{100} \mathrm{~m} / \mathrm{s}$

MHT CET 2019 2nd May Evening Shift

MCQ (Single Correct Answer)

-0

Three identical rods each of mass ' $M$ ' and length ' $L$ ' are joined to form a symbol ' $H$. The moment of inertia of the system about one of the sides of ' $H$ ' is

$\frac{2 M L^2}{3}$

$\frac{M L^2}{2}$

$\frac{M L^2}{6}$

$\frac{4 M L^2}{3}$

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