Elasticity Numerical on Stress Strain Young’s Module Class 11 Physics ISC Nootan Solutions Ch-13. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Elasticity Numerical on Stress Strain Young’s Module Class 11 Physics ISC Nootan Solutions
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Writer | Kumar and Mittal |
| Publication | Nageen Prakashan |
| Chapter-13 | Elasticity |
| Topics | Numericals on Elasticity, Stress, Strain and Young’s Modulus |
| Academic Session | 2024-2025 |
Numericals on Elasticity, Stress, Strain and Young’s Modulus
Class 11 Physics ISC Nootan Solutions Ch-13 of Kumar and Mittal Physics , Nageen Prakashan
Que-1: A weight of 1.0 kg is suspended from the lower end of a wire of cross-section 10 mm^2. Find the magnitude and direction of the stress produced in it. (g = 9.8 m/s^2)
Solution- Stress produced is given by,
T = F/A
T = 9.8/(10^-5)
T = 9.8 × 10^5 N/m^2 Ans.
Stress produced is 9.8 × 10^5 N/m^2 in vertically upward direction.
Que-2: In order to produce a longitudinal strain of 2 x 10^-4, a stress of 2.4 x 10^7 N/m^2 is produced in a wire. Calculate the Young’s modulus of the material of the wire.
Solution- by hooke’s law
stress = young’s modulus * longitudinal strain
young’s modulus = stress / strain
young’s modulus = (2.4 x 10^7)/(2 x 10^-4)
= 1.2 x 10^11 N/m^2 Ans.
Que-3: A copper wire has a length of 2.2 m. Its diameter is 7 x 10^-4 m. What will be the elongation in its length when a 5 kg-weight is suspended?(Young’s modulus of copper, Y = 1.25×10^11 N/m^2 and g = 9.8 m/s^2)
Solution
Force of tension on the wire
F = mg
F = 5 x 10
F = 50N
Elongation is given by
ΔL = FL/AY
ΔL = FL/(π/4)(7×10^-4)² x 1.25 x 10¹¹
ΔL = 0.002286 m = 0.0023 m
= 2.3 mm
The elongation of the wire is 2.3 mm. Ans.
Que-4: If Young’s modulus for steel is 2 x 10^11 N/m^2, then how much weight be suspended from a steel wire of length 2.0 m and diameter 1.0 mm so that the length of a wire be increased by 1.0 mm? (g = 9.8 m/s^2)
Solution- Y = 2.0 x 10^11 N/m^2
d = 1 mm
r = 0.5 mm = 0.5 x 10^-3 m
l = 1 mm = 1 x 10^-3 m
According to formula
Y = MgL/πr^2l
M = (πr^2l x Y)/gL
= {3.14 x (0.5 x 10^-3)^2 x 1 x 10^-3 x 2.0 x 10^11}/ 9.8 x 2
= 8 kg Ans.
Que-5: If the length of a wire increased by 1 mm under 1 kg weight , what will be the increase under 2 kg? Under 100 kg?
Solution –
k = F/e
k = spring constant
F = force = mg
e = Extension.
Take g = 10
F = 1 × 10 = 10 N
k = 10/1 = 10
Now for 2kg :
F = 2 × 10 = 20 N
e = F/k
hence :
20/10 = 2mm Ans
Wire will break as the stretching force is beyond the breaking point of wire.
Que-6: The length of a wire increased by 1% on loading a 2 kg weight on it. Calculate the linear strain in the wire.
Solution- Strain = change in length / original length
= (1% of L)/L
= (L/100)L
= 0.01 Ans.
Que-7: The length of a wire increases by 8 mm when a weight of 5 kg is suspended from it. If other things remain the same but the radius of the wire is doubled, what will be the increase in its length?
Solution- l = FL/πr^2 Y
When the radius is doubled, let l. be the increase in length.
l. = FL/π(2r^2)Y
l./l = 1/4
l.= l/4
l. = 8/4
l.= 2 mm Ans.
Que-8: If Young’s modulus of steel is 2.0 x 10^11 N/m^2, then how much weight be suspended from the steel wire of length 2.0 m and diameter 1.0 mm so that the length of the wire be increased by 1.0 mm
Solution- Y = 2.0 x 10^11 N/m^2
d = 1 mm
r = 0.5 mm = 0.5 x 10^-3 m
l = 1 mm = 1 x 10^-3 m
According to formula
Y = MgL/πr^2l
M = (πr^2l x Y)/gL
= {3.14 x (0.5 x 10^-3)^2 x 1 x 10^-3 x 2.0 x 10^11}/ 9.8 x 2
= 8 kg Ans.
Que-9: A composite wire consists of a steel wire of length 1.5 m and a copper wire of length 2.0 m, with a uniform cross-sectional area of 2.5 x 10^-5 m^2. It is loaded with a mass of 200 kg. Find the extension produced. Young’s modulus of copper is 1.0 x 10^11 N/m^2 and that of steel is 2.0 x 10^11 N/m^2. Take g = 9.8 m/s^2.
Solution- l = l of steel + l of copper
(MgLs)/(πr^2 Ys) + (MgLc)/(πr^2 Yc)
Mg/πr^2 {Ls/Ys + Lc/Yc}
= {(200 x 9.8)/(3.14 x 2.5 x 10^-5)} {(1.5/2.0x 10^11) + (2/1.0 x 10^11)}
= 2.156 x 10^-3 m
= 2.156 mm Ans.
Que-10: A rod AD consisting of three segments AB, BC and CD joined together, is hanging vertically from a fixed support A. The lengths of a segments are respectively 0.1 m, 0.2 m and 0.15 m. The cross-section of the rod is uniformly 10^-4 m^2. A weight of 10 kg is hung from D. Calculate the displacements of the points B, C and D. Neglect the weight of the rod. Given, YAB = 2.5 x 10^10 N/m^2 , YBC = 4.0 x 10^10 N/m^2 and YCD = 1.0 x 10^10 N/m^2.
Solution- Given:
A = 10^−4 m^2
YAB = 2.5×10^10 N/m^2
YBC = 4×10^10 N/m^2
YCD = 1×10^10 N/m^2
As we know,
ΔL=FL/AY=mgL/AY = (10 × 10L)/(10^−4 Y) = 10^6 L/Y
AB : (ΔL)1 = 10^6 × LAB/YAB
= 10^6 × 0.1/(2.5×10^10)
= 4×10^−6 m
BC : (ΔL)2 = 10^6 x LBC/YBC
= 10^6 × 0.2/(4×10^10)
= 5×10^−6 m
CD : (ΔL)3 = 10^6 × LCD/YCD
= 10^6 × 0.15/(1×10^10)
= 15×10^−6 m
Displacement of D:
ΔL1+ΔL2+ΔL3 = 23.52×10^−6m Ans.
Que-11: A piece of metal of 2 kg weight is suspended from one end of a vertical wire whose other end is fixed and the piece is fully immersed in an oil of density 0.7 x 10^3 kg/m^3. The length of the wire increases by 1 mm. If the diameter of the wire is 0.6 mm, Young’s modulus is 2.0 x 10^11 N/m^2 and the volume of the the metal-piece is 800 cm^3, then calculate the initial length of the wire.
Solution- Given that,
Mass of piece m = 2kg
Oil density p = 0.7 x 10³ kg/m³
Increases length Δd = 1mm
Diameter of wire D = 0.6mm
Young’s modulus Y = 2 x 10¹¹ N/m²
Volume of metal piece V = 800 cm³
According to buoyancy force
If the metal piece fully emerged then the force,
F = mg – pgV
Now, the formula of Young’s modulus
Y = (F x d)/(A x Δd)
Now, put the value of F in Young’s modulus
d = (Y x A x Δd)/(mg – pgV)
d =(2 x 10¹¹ x 3.14 x 0.3 x 0.3 x 10^-6 x 1 x 10¯³)/{(2 x 9.8) – (0.7 x 10³ x 9.8 x 800 x 10^-6)}
d = 0.562 x 10²/14.112 m
d = 0.04005 x 10² m
d = 4m Ans.
Que-12: A weight of 20 kg is suspended from a wire. The area of cross-section of the wire is 1 mm^2 and when stretched, the length is exactly 6 m. When the weight is removed, the length reduces to 5.995 m. Calculate the Young’s modulus (Y) of the material of the wire.
Solution- Given that,
Weight = 20 kg
Area of cross section = 1 mm²
Stretched length = 6 m
Removed length = 5.995 m
We need to calculate the Young’s modulus of the wire
Using formula of young’s modulus
Y = stress/strain
Y = mg/A x l/Δl
Put the value into the formula
Y = (20/1 x 10^-6) x (5.995/5 x 10^-3)
Y = 2.398 x 10^10 N/m² Ans.
Que-13: The length of a rubber cord increases to 60 cm under a weight of 100 g suspended from it and to 70 cm under a weight of 120 g. Find the initial length of the cord. Also calculate the weight under which the length of the cord will become 74 cm.
Solution- k = spring constant
L₀ = initial length of cord
L₁ = length of cord when 100 g weight is suspended = 60 cm
m = 100 g
using equilibrium of force
k (L₁ – L₀) = mg eq-1
L₂ = length of cord when 120 g weight is suspended = 70 cm
M = 120 g
using equilibrium of force
k (L₂ – L₀) = Mg eq-2
dividing eq-1 by eq-2
k (L₁ – L₀) /(k (L₂ – L₀)) = mg/(Mg)
(60 – L₀)/(70 – L₀) = 100/120
L₀ = 10 cm Ans.
By using eq-1
k (L₁ – L₀) = mg
k (60 – 10) (0.01) = (0.1) (9.8)
k = 1.96 N/m
let the weight hanged be “m’ ”
L₂ = 74 cm
using equilibrium of force
k (L₂ – L₀) = m’ g
(1.96) (0.74 – 0.1) = m’ (9.8)
m’ = 0.128 kg
m’ = 0.128 x 1000 g
m’ = 128 g Ans.
Que-14: A substance breaks down by a stress of 10^9 N/m^2. If the density of the substance be 3 x 10^3 kg/m^3, find that length of the wire made of the same substance, by which it will break under its own weight when suspended.
Solution- Stress = F/A = Mg/A
= (A x L x p x g)/A
10^9 = Lpg
L = 10^9/pg
L = 10^9/3 x 10^3 x 9.8
L = 3.4 x 10^4 m Ans.
Que-15: For steel the breaking stress is 8.0 x 10^6 N/m^2 and the density is 8.0 x 10^3 kg/m^3. Find the maximum length of a steel wire which can be suspended without breaking under its own weight.(g = 9.8 m/s^2)
Solution- Weight = Mg = Vdg = Aldg
Breaking stress = Mg/A = Aldg/A = Ldg
[L = maximum length]
∴L = Breaking stress/dg
= (8×10^6)/(8×10^3×10)
= 100 m Ans.
Que-16: A body of 10 kg tied at one end of a 0.3 m wire is rotated in a horizontal circle. The area of cross-section of the wire is 10^-6 m^2 and the breaking-stress is 4.8 x 10^7 N/m^2. By how much maximum angular speed can the body be rotated?
Solution- Given: length, l = 0.3 m
Mass of the body, m = 10 kg
Breaking stress, σ = 4.8×10^7 N/m^2
Area of cross-section, a = 10^−6 m^2
Maximum angular speed ω=?
Now, T=mlω^2
And, σ = T/A = mlω^2/A
⇒ mlω^2/A ≤ 4.8×10^7
⇒ ω^2 ≤ {(4.8×10^7)A}/ml
⇒ ω^2 ≤ {(4.8×10^7)(10^−6)}/10×0.3=16
⇒ ω = 4 rad/s Ans.
Que-17: A metallic wire has a radius of 0.2 mm. How much force is required to have an increase of 0.2% in its length? (Y = 9.0 x 10^10 N/m^2).
Solution- change in length, ∆l = 0.2% of length of wire
= 0.2% of l
= 0.2/100 × l
= 0.002l
so strain = ∆l/l = 0.002l/l = 0.002
cross sectional area, A = πr²
= 3.14 × (0.2mm)²
= 3.14 × (2 × 10^-4m)²
= 3.14 × 4 × 10^-8 m²
= 12.56 × 10^-8 m²
now Young’s modulus = stress/strain
⇒Y = (F/A)/strain
⇒9 × 10^10 = F/(12.56 × 10^-8 × 0.002)
⇒F = 9 × 10^10 × 12.56 × 10^-8 × 0.002
= 9 × 12.56 × 0.2
= 22.6 N Ans.
Que-18: If a stress of 1 kg-wt/mm^2 is applied on a wire, then what will be the percentage increase in the length of the wire? (Y = 1.0 x 10^11 N/m^2 and 1 kg-wt = 9.8 N).
Solution- l/L = F/Ay
= (1×9.8)/(10z^−6 × 10^11)
= (10^−5 × 9.8 × 100)%
= 0.0098% Ans
Que-19: A wire of length 1.0 m has area of cross-section 0.1 cm^. The Young’s modulus of material of the wire is 2.5 x 10^11 N/m^2 and coefficient of linear thermal expansion is 2 x 10^-5 /°C. The wire is clamped between two rigid supports and its temperature is decreased by 10°C. Calculate the tension developed in it.
Solution- T = YA ∝ Δt
= 2.5 x 0.1 x10^-4 x 2 x 10^5 x 10
= 500 N Ans.
Que-20: The area of cross-section of a metal-rod is 1 cm^2. It is clamped tightly at its two ends at 0°C. The rod is heated to 100°C. Calculate the tension produced at its two ends. Coefficient of thermal expansion of metal is 1.2 x 10^-5/°C and Young’s modulus is 2.0 x 10^11 N/m^2.
Solution – T = YA ∝ Δt
= 2 x 10^11 x 1 x 10^-4 x 1.2 x 10^-5 x 100
= 2.4 x 10^4 N Ans.
—: end of Elasticity Numerical on Stress Strain Young’s Module Class 11 ISC Nootan Solutions Ch-13 :—
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