Properties of Matter (Physics)
Multiple Choice Questions
1. Consider two wires X and Y. The radius of wire X is 3 times the radius of Y. If they are stretched by the same load then the stress on Y is
(a) equal to that on X
(b) thrice that on X
(c) nine times that on X
(d) Half that on X
Answer : c) nine times that on X
Solution:
σx ∝ 1 / rx2 ;
σy ∝ 1 / ry2
σx / σy = (ry / rx )2 = (ry / 3ry )2
σx / σy = (1/3)2 = 1 / 9
σy = 9 σx
2. If a wire is stretched to double of its original length, then the strain in the wire is
(a) 1
(b) 2
(c) 3
(d) 4
Answer : a) 1
Solution:
ε = Δl / l
ε = [ lt – lo ] / lo
= (2 – 1) / 1 = 1 / 1 = 1
3. The load – elongation graph of three wires of the same material are shown in figure. Which of the following wire is the thickest?
(a) wire 1
(b) wire 2
(c) wire 3
(d) all of them have same thickness
Answer : a) wire 1
Solution:
If slope is high the elongation will be less
4. For a given material, the rigidity modulus is (1/3) rd of Young’s modulus. Its Poisson’s ratio is
(a) 0
(b) 0.25
(c) 0.3
(d) 0.5
Answer : d) 0.5
Solution:
η = 1/3 y
y = 2 η (1 + μ )
y = 2y/3 (1 + μ )
3/2 = 1 + μ
μ = (3/2) – 1 = 1 / 2
μ = 0.5
5. A small sphere of radius 2cm falls from rest in a viscous liquid. Heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to
(a) 22
(b) 23
(c) 24
(d) 25
Answer : d) 25
Solution:
dQ / dt = E / t = Power
Power = P = F × vt
F = 6πηrvt
dQ / dt = 6πηrvt2
vt = [2/9] [ (ρ – σ ) / η]g × [ r2]
dQ / dt = 6πηr × ( 2/9 [ (ρ–σ)/η × g ])2 × r4
dQ / dt ∝ r5
6. Two wires are made of the same material and have the same volume. The area of cross sections of the first and the second wires are A and 2A respectively. If the length of the first wire is increased by ∆l on applying a force F, how much force is needed to stretch the second wire by the same amount?
(a) 2
(b) 4
(c) 8
(d) 16
Answer : b) 4 F
Solution:
y = Fl / [Δl × A]
y = [ Fl / AΔl ] × [ A / A ] = [F.V ] / [Δl × A2]
F ∝ A2
F2 / F1 = [ (2A)2 ] / [A2 ] = 4
F2 = 4 F1
7. With an increase in temperature, the viscosity of liquid and gas, respectively will
(a) increase and increase
(b) increase and decrease
(c) decrease and increase
(d) decrease and decrease
Answer : c) decrease and increase
8. The Young’s modulus for a perfect rigid body is
(a) 0
(b) 1
(c) 0.5
(d) infinity
Answer : d) infinity
Solution:
Y = Fl / ΔlA = Fl / [O × A] = Fl / O
Y = ∞
9. Which of the following is not a scalar?
(a) viscosity
(b) surface tension
(c) pressure
(d) stress
Answer : d) stress
10. If the temperature of the wire is increased, then the Young’s modulus will
(a) remain the same
(b) decrease
(c) increase rapidly
(d) increase by very a small amount
Answer: b) decrease
Solution:
Y = (Fl) / (A.Δt)
⇒ Y ∝ 1 / Δt
11. Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is ∆l. If Y represents the Young’s modulus, then which of the following graphs is a straight line?
(a) ∆l verses V
(b) ∆l verses Y
(c) ∆l verses F
(d) ∆l verses 1/l
Answer: c) Δl versus F
Solution:
Y = (Fl ) / (A. Δl) = Fl / V
12. A certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then
(d) energy is neither released nor absorbed
Answer : c) energy = 3VT [ 1/r – 1/R ] is released
13. The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
(a) length = 200 cm, diameter = 0.5 mm
(b) length= 200 cm, diameter = 1 mm
(c) length = 200 cm, diameter = 2 mm
(d) length= 200 cm, diameter = 3 m
Answer : a) length = 200 cm, diameter = 0.5 mm
Solution:
Δl = Fl / AY = Fl / πr2Y
Δl ∝ l / r2 ,
Δl ∝ l / d2
14. The wettability of a surface by a liquid depends primarily on
(a) viscosity
(b) surface tension
(c) density
(d) angle of contact between the surface and the liquid
Answer : d) angle of contact between the surface and the liquid
15. In a horizontal pipe of non-uniform cross section, water flows with a velocity of 1 m s-1 at a point where the diameter of the pipe is 20 cm. The velocity of water (m s-1) at a point where the diameter of the pipe is
(a) 8
(b) 16
(c) 24
(d) 32
Answer : b) 16
Solution:
a1v1 = a2v2
π [d12v1 / 4] = π [d22v2 / 4] . v2
d22 = [v1 / v2 ] d12
= (1/1.5) × 400 × 10-4
= 2.66 × 10-4
d2 = 16.3 × 10-2
= 16.3 cm
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