Mechanical Properties Of Solids(12th Grade > Physics ) Questions and Answers
Explanation:-
Answer: Option D. -> α1l1=α2l2:
D
d)L2=l2(l+α2△θ) and L1=l1(l+α1△θ)
⇒(L2−L1)=(l2−l1)+△θ(l2α2−l1α1)
Now (L2−L1)=(l2−l1) so, l2α2−l1α1=0
Explanation:-
Answer: Option C. -> length of rod:
C
Lt=L0(1+α△θ)
△L=Lt−L0=L0α△θ
If the same rod of lenghth L0 is subjected to stress along its length, then extension in length can be calculated by Hooke's law
Y=stressstrain=stress△LLn
=L0×stress△L
Therefore △L=L0×stressY
If the rod is prevented from expanding, we have
Loα△θ=Lo×stressY
Therefore stress=Yα△θ= (independent of L0)
Explanation:-
Answer: Option D. -> 4%:
D
Given Poisson's ration= 0.5
It shows that the density of material is constant
Therefore the change in volume of the wire is zero
Thus
V=A×l =constant
log V=log A+log |
therefore △VV=0=△AA+△ll
△ll=−△AA
or % increase in length=△ll×100
=-(-4)
=4%
Explanation:-
Answer: Option C. -> more than that of rubber:
C
Iron is more elastic than rubber, hence the number of atoms per unit volume will be more in case of iron.
Question 8. A cube is compressed at 0∘ C equally from all sides by an external pressure p. By what amount should the temperature be raised to bring it back to the size it had before the external pressure was applied? (Given K is bulk modulus of elasticity of the material of the cube and α is the coefficient of linear expansion)
Explanation:-
Answer: Option B. -> p3Kα:
B
k=pV△v=pVy△T=p3αT
or T=p3kα
Explanation:-
Answer: Option A. -> 5.2 mm:
A
The contraction in the length of the wire due to change in the temperature
=αLΔT
=1.2×10−5×3×(103−303)
=−7.2×10−3m
The expansion in the length of wire due to stretching force
=FLAY=(10X10)X3(0.75×10−6)(2×1011)
=2×10−3m
Resultant change in length
=−7.2×10−3+2×10−3
=−5.2×10−3
=-5.2 mm
Negative sign shows contraction
Question 10. Two rods of different materials having coefficients of linear expansion α1 and α2 and Young’s modulus
Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If
α1 : α2 = 2:3, the thermal stress developed in the two rods are equal provided Y1:Y2 equal to
Y1 and Y2 respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If
α1 : α2 = 2:3, the thermal stress developed in the two rods are equal provided Y1:Y2 equal to
Explanation:-
Answer: Option D. -> 3 : 2:
D
From the formula for thermal stress
(F/A)1=α1y1△T
(F/A)2=α2y2△T
Since,temperature △T is same
(FA)1(FA)2=α1y1α2y2
For thermal stress to be equal
α1y1=α2y2
ory1y2=α2α1=32 since α1α2=2:3)