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XI DPP-26 CIRCULAR MOTION

Authored by Target Physics

Physics

11th - 12th Grade

Used 1+ times

XI DPP-26 CIRCULAR MOTION
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20 questions

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1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A rubber band of length l has a stone of mass m tied to its one end. It is whirled with speed v so that the stone describes a horizontal circular path. The tension T in the rubber band is -

zero

mv2 /l

> (mv2)/l

< mv2 /l

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

A smooth wire is bent into a vertical circle of radius a . A bead P can slide smoothly on the wire. the circle is rotated about diameter AB as axis with a speed  ω\omega  as shown in figure. The bead P is at rest with respect to the circular ring in the position shown. Then  ω2\omega^2 is equal to-

2g/a

 2g(a3)\frac{2g}{\left(\frac{a}{\sqrt{3}}\right)}  

 g3ga\frac{g\sqrt{3}g}{a}  

 2a(g3)\frac{2a}{\left(\frac{g}{\sqrt{3}}\right)}  

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A simple pendulum of length L and mass M is oscillating in a plane about a vertical line between angular limits – ϕ and + ϕ. For an angular displacement ϕ [ | θ | < ϕ], the tension in the string and the velocity of the bob are T and v respectively. The following relation holds good under the above conditions-

T = Mg cos θ

T cos θ = Mg

T – Mg cos θ = Mv2/L

T + Mg cos θ = Mv2/L

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A car is moving with a speed of 30 m/sec on a circular path of radius 500 m. Its is increasing at the rate of 2 m/sec2. What is the acceleration of the car ?

9.8 m/sec2

2.7 m/sec2

2.4 m/sec2

1.8 m/sec2

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A point moves along a circle with velocity v = at where a = 0.5 m/sec2. Then the total acceleration of the point at the moment when it covered (1/10) th of the circle after beginning of motion -

0.5 m/sec2

0.6 m/sec2

0.7 m/sec2

0.8 m/sec2

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A solid body rotates about a stationary axis so that its angular velocity depends on the rotation angle  ϕ\phi   as  ω=ω0kϕ\omega=\omega_0-k\phi  , where  ω0\omega_0   and k are positive constants. At the moment t = 0, the angle ϕ\phi   = 0. Find the time dependence of rotation angle -

K.ω0ektK.\omega_0e^{-kt}

ω0K[ekt]\frac{\omega_0}{K}\left[e^{-kt}\right]

ω0K[1ek.t]\frac{\omega_0}{K}\left[1-e^{-k.t}\right]

Kω0[ekt1]\frac{K}{\omega_0}\left[e^{-kt}-1\right]

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed (gl)\sqrt{\left(gl\right)}   . Then the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string equal the weight of the particle-

 3lg,cos1(32)\sqrt{\frac{3l}{g}},\cos^{-1}\left(\frac{3}{2}\right)  

 l g3,cos1(23)\sqrt{\frac{l\ g}{3}},\cos^{-1}\left(\frac{2}{3}\right)  

 3gl,cos1(23)\sqrt{\frac{3g}{l}},\cos^{-1}\left(\frac{2}{3}\right)  

 gl3, sin1(23)\sqrt{\frac{gl}{3}},\ \sin^{-1}\left(\frac{2}{3}\right)  

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