51 - Mechanics Part 2 Questions Answers

Physics >> Mechanics Part 2 >> Elasticity Medical Exam

A nylon rope 3 cm in diameter has a breaking strength of 1.5 x 105 N. The breaking strength for a similar rope 1.5cm in diameter is

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Joshi sir comment

breaking stress will remain same

Physics >> Mechanics Part 2 >> Elasticity Medical Exam

A cubical block of a metal having shear modulus η is fixed from one face as shown in figure and constant force F is applied on it. Find the shear strain produced in it.

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Joshi sir comment
Shear elasticity = F/A/shear strain Now solve
Physics >> Mechanics Part 2 >> Gravitation Medical Exam

If L is the angular momentum of a satellite revolving around earth is a circular orbit of radius r with speed v, then (i) L α v

(ii) L α r

(iii) L α √v

(iv) L α √r

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Joshi sir comment

L = m*v*r

 

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at height h above the surface of earth is [ R is radius of earth and M is mass of earth]

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[-GMm/(R+h)]-[-GMm/R] = GMm[1/R - 1/(R+h)]

solve ?

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

Two points masses having m and 4m are placed at distance at r. The gravitational potential at a point, where gravitational field intensity zero is

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Joshi sir comment

field intensity at distance x from m = Gm/x2 - G4m/(r-x)2

compare it to 0 and find x

then calculate gravitational potential at the point 

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

During motion of a planet from perihelion to aphelion the work done by gravitational force of sun on it is positive or negative.

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Joshi sir comment

change in K E = work done by all the forces

and from perihelion to aphelion velocity decreases

so work done will be negative

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

If an object is projected vertically upwards with speed, half the escape speed of earth , then the maximum height attained by it is [R is radius of earth]

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Joshi sir comment

from ground to that point 

1/2 m[√(2gR)/2] + [-GMm/R]  =  0 + [-GMm/(R+h)]

solve for h

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

Two point masses having mass m and 2m are placed at distance d. The point on the line joining point masses, where gravitational field intensity is zero will be at distance

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Joshi sir comment

let the point will be at a distance of x from m then compare the gravitational field of the two masses at that point

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

If earth suddenly stop rotating, then the weight of an object of mass m at equator will [ ω is angular speed of earth and radius R is its radius]

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Joshi sir comment

In rotating condition mass at equator is mg-mRω2

if earth stops rotation then it will become mg so increases by mRω2

Physics >> Mechanics Part 2 >> Gravitation Medical Exam

Three particles A,B and C each of mass m are lying at the corners of an equilateral triangle of side L. If the particle A is released keeping the particles B and C fixed, the magnitude of instantaneous acceleration of A is

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Joshi sir comment

force by the other two on A = Gmm/L2 each but the angle between the two forces is 60 degree so resultant force = √3Gmm/L2

now calculate acceleration?

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