575 - Physics Questions Answers

Physics >> Mechanics Part 2 >> Elasticity Medical Exam

The work done per unit volume to stretch the length of area of cross-section 2 mm2 by 2% will be [ Y = 8 x 1010 N/m2 ]  

Asked By: SWATI KAPOOR
is this question helpfull: 7 13 submit your answer
Joshi sir comment
W =1/2 Y (dl/l)2 V V represents volume Now solve
Physics >> Mechanics Part 2 >> Elasticity Medical Exam

To break a wire a stress of 9 x 105 N/m2 is required. If density of the material of wire is 3 g/cc, then the minimum length of wire which will break by its own weight will be 

Asked By: SWATI KAPOOR
is this question helpfull: 2 0 submit your answer
Joshi sir comment

use stress = force /area

Physics >> Mechanics Part 2 >> Elasticity Medical Exam

A steel wire is 1m long and 1mm2 in area of cross-section. If it takes 200 N to stretch this wire by 1mm, how much force will be required to stretch a wire of the same material as well as diameter from its normal length of 10m to a length of 1002 cm?

Asked By: SWATI KAPOOR
is this question helpfull: 4 0 submit your answer
Joshi sir comment

by first set of data calculate Y then by using this Y solve the question

Physics >> Mechanics Part 2 >> Elasticity Medical Exam

The length of wire, when M1 is hung from it, is l1 and is l2 with both M1 and M2 hanging. The natural length of wire is [ M2 is on M1 ]

Asked By: SWATI KAPOOR
is this question helpfull: 1 0 submit your answer
Joshi sir comment

Y = M1gL/A(l1-L) = (M1+M2)gL/(l2-L)

solve for L

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

Asked By: SWATI KAPOOR
is this question helpfull: 1 0 submit your answer
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.

Asked By: SWATI KAPOOR
is this question helpfull: 0 4 submit your answer
Joshi sir comment
Shear elasticity = F/A/shear strain Now solve

a package of mass 10kg is released from rest on a rough slope inclined at 25 degrees to the horizontal. after 2 seconds the package has moved 4m down the slope. find the coefficient of friction between the package and the slope.

 

a box of mass 2kg is pushed up a rough plane by a horizontal force of magnitude 25N. the plane is inclined to the horizontal at an angle of 10 degree. given that the coefficient of friction between the box and the plane is 0.3, find the acceleration of the box.

Asked By: AMAR KHAN
is this question helpfull: 7 1 read solutions ( 1 ) | submit your answer
Joshi sir comment

first part is correctly explained by Sarika

for second part make components of forces as 25 N and weight along and perpendicular to the incline, friction μN will be along downward incline and N will be the resultant of components of weight and 25 N perpendicular to the plane. 

In order to raise a mass of 100 kg, a man of mass 60 kg fastens a rope to it and passes the rope over a smooth pulley. He climbes the rope with an acceleration 5g/4 relative to the rope. The tension of the rope is:

(a) 1432 N

(b) 928 N

(c) 1218 N

(d) 642 N

Asked By: RAJIV
is this question helpfull: 7 1 submit your answer
Joshi sir comment

eq. for upward motion of 100 Kg is T-100g = 100a  (1)

for man eq. is T-60g = 60[(5g/4)-a]             from ground frame

solve?

A gas expands against a variable external pressure given by p = 10/V atm, where V is the volume at each stage of expansion. in expanding from 10 L to 100 L, the gas undergoes a change in internal energy ΔU = 418 J. How much heat has been absorbed?

Asked By: RICHIK BANDYOPADHYAY
is this question helpfull: 3 4 read solutions ( 1 ) | submit your answer
Joshi sir comment

process is isothermal because pV = 10 = nRT

so work done W = 2.303nRTlog(V2/V1)

so Q = Δ+ W

 

if an area of isoscale right angle triangle(made of uniform wire of unit resistivity) is 8 unit. a battery of emf 10 unit is connected with across the longest branch then the current drawn from battery is approx                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     ans is  3A                                                                                                   

Asked By: AKANKSHA SINHA
is this question helpfull: 5 2 submit your answer
Joshi sir comment

area = l2/2 = 8 so l = 4

ρ = 1 

so  resistance of the 3 legs of the triangle = 4/A, 4/A, 4√2/A

battery is connected across the longest branch so eq. resistance = 4/A+4/A+4√2/A = 13.6/A

i think area of cross section will be given then calculate current

Login Here

Register