Classical Mechanics (CSIR Physical Sciences): Questions 53 - 56 of 59

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Question number: 53

» Classical Mechanics » Special Theory of Relativity » Mass-Energy Equivalence

Appeared in Year: 2011

MCQ▾

Question

An atom of mass M can be excited to a state of mass M+Δ by photon capture. The frequency of a photon which can cause this transition is – (December)

Choices

Choice (4) Response

a.

Δ(Δ+2M)c22Mh

b.

Δc22h

c.

Δc2h

d.

Δ2c22Mh

Question number: 54

» Classical Mechanics » Special Theory of Relativity » Relativistic Kinematics

Appeared in Year: 2011

MCQ▾

Question

A planet of mass m moves in the inverse square central force field of the Sun of mass M. If the semi – major and semi – minor axes of the orbit are a and b , respectively, the total energy of the planet is: (December)

Choices

Choice (4) Response

a.

GMm(1a+1b)

b.

GMma+b

c.

GMmab(a+b)2

d.

GMma(1b1a)

Passage

In the absence of an applied torque a rigid body with three distances principal moments of inertia given by I1,I2andI3 is rotating freely about a fixed point inside the body. The Euler equations for the components of its angular velocity (ω1,ω2,ω3) are

ω˙1=I2I1I1ω2ω3,ω˙2=(I3I1)I2ω1ω3,ω˙3=I1I2I3ω1ω2 (June)

Question number: 55 (1 of 1 Based on Passage) Show Passage

» Classical Mechanics » Rigid Body Dynamics - Moment of Inertia Tensor

Appeared in Year: 2011

MCQ▾

Question

The constant of motion are –

Choices

Choice (4) Response

a.

I1ω12+I2ω22+I3ω32andω1+ω2+ω3

b.

I1ω12+I2ω22+I3ω32andI12ω12+I22ω22+I32ω32

c.

ω12+ω22+ω32andI1ω1+I2ω2+I3ω3

d.

ω12+ω22+ω32andI12ω12+I22ω22+I32ω32

Question number: 56

» Classical Mechanics » Special Theory of Relativity » Lorentz Transformations

Appeared in Year: 2016

MCQ▾

Question

Let (x,t) and (x,t) be the coordinate systems used by the observers O and O , respectively. Observer O moves with a velocity v=βc along their common positive x axis. If x+=x+ct and x=xct are the linear combinations of the coordinates the Lorentz transformation relating O and O takes the form – (June)

Choices

Choice (4) Response

a.

x+=1+β1βx+andx=1β1+βx

b.

x+=1β1+βx+andx=1+β1βx

c.

x+=xβx+1β2andx=x+βx1β2

d.

x+=x+βx1β2andx=xβx+1β2

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