Elasticity Numerical on Bulk Modulus Rigidity and Elastic Potential Energy Class 11 ISC Physics 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 Bulk Modulus Rigidity and Elastic Potential Energy Class 11 ISC Physics Nootan Solutions Ch-13

Board | ISC |

Class | 11 |

Subject | Physics |

Writer | Kumar and Mittal |

Publication | Nageen Prakashan |

Chapter-13 | Elasticity |

Topics | Numericals on Bulk Modulus Rigidity and Elastic Potential Energy |

Academic Session | 2024-2025 |

### Numericals on Bulk Modulus Rigidity and Elastic Potential Energy

**Que-21: A brass rod of length 0.2 m and area of cross-section 2 cm² is compressed by a force of 5 kg-wt along its length. Determine the increase in the energy of the rod. Young’s modulus of brass is 1.4 x** **10^¹¹ N/m². **

**Solution-** Energy = 1/2 Fl

= 1/2 X F X (FL/AY)

= 1/2 X F^2 L/AY

= 1/2 X {(50)^2 X 20 X 10^−2}/ {2x 10^−4 X 1.4 X 10^11

= 8.57 X 10^−6 J Ans.

**Que-22: A fluid of volume 1.5 x 10‾³ m³ when subjected to a pressure-change of 5.0 x 10^6 N/m****² has its volume reduced by 3.0 x 10^-7 ****m³. Find the coefficient of bulk modulus of the fluid.**

**Solution- **Given pressure change ◇P= p= 5.0×10^6 N/m^2

volume of fluid V = 1.5 × 10^3 m^3

change in volume ΔV = v =3.0 × 10^-7 m^3

volume strain = v/V = 3.0 × 10^-7/1.5 × 10^-3

= 2.0 × 10^-4

Bulk modulus, B = change in pressure/volume strain

=P/(v/V)

= 5.0 × 10^6/ 2.0 × 10^-4

= 2.5 × 10^10 N/m^2 Ans.

**Que-23: 1 ****m³ of water is taken from the surface of a lake to a depth of 200 m inside the lake. What will be the change in its volume, if the bulk modulus of elasticity of water is 22000 atmospheres? (Density of water = 1 x 10****³ kg/m****³, atmospheric pressure = 10^5 N/m****² and g = 9.8 m/s****²****).**

**Solution-** Bulk modulus is defined as

K = P/(ΔV/V)

∴ΔV = PV/K

Here

P = hρg = 200 × 10^3 × 10 N/m^2

K = 22000 atm = 22000 × 10^5 N/m2

V = 1 m^3

ΔV = (200 × 10^3 × 10 × 1)/ 22000 × 10^5

= 9.1×10^−4 m^3 Ans

**Que-24: Find the change in volume which 1 m³ of water will undergo when taken from the surface, to the bottom of sea 1 km deep. Given bulk modulus of elasticity of water is 20,000 atmosphere and 1 atmosphere = 1.013 x 10^5 N/m².**

**Solution-** p = 100 m of water column

= (100×100) × 1 × 980dyne/sq.cm ΔV=?

B = 22000 atm. = 22000 × 76 × 13.6 × 980dyne/sq.cm.

Now, ΔV = pV/B

= (100 ×100×980×1)/(22000 × 76 × 13.6 × 980)

= 4.83 10^-3 m^3 Ans.

**Que-25: A cube is subjected to pressure of 5 x 10^5 N/m². Each side of the shortened by 1%. Find volume strain and bulk modulus of elasticity of cube.**

**Solution-** Let l be the initial length of each side of cube.

Final length = (1−(1/100)l) = (99/100)l

Initial volume Vi = t^3

Final volume , Vf = {(99/100) l}^3 − l^3

Volumetric strain = (ΔV) = Vf − Vi

= (1−1/100)^3 − 1

= −3/100 = −0.03 Ans.

Normal stress = increase in pressure

= 5 × 10^5 N/m^2

∴ Bulk modulus of elasticity , B = (5 × 10^5)/ −0.03

= 1.67×10^7 N/m^2 Ans.

**Que-26: A metal cube of side 10 cm is subjected to a shearing stress of 10^4 N/m². Calculate the modulus of rigidity if the top of the cube is displaced by 0.05 cm with respect to its bottom.**

**Solution-** We know that shearing stress = F/A = 10^4 N/m^2

Length of side of cube = 10cm = 10/100 m = 0.1m

Shearing displacement = △n = 0.05cm

= 0.05/100 m = 0.0005m

We know that modulus of rigidity is η = F/Aθ

= FL/A △x [tanθ≈θ=△x/L]

= (10^4 × 0.1)/(5×10^−4)

⇒ 10^7/5^1

⇒ 2 × 10^6 N/m^2 Ans.

**Que-27: The Young’s modulus of brass is 10^11 N/m^2. Find increase in energy of rod of brass of cross-sectional area 10^-4 m^2 and length 0.1 m when it is compressed along its length with a force of 10 kg-wt.**

**Solution-** Increase in energy

ΔE=1/2× stress × strain × volume

ΔE=1/2×F/A×F/(A×Y)×(A×L)

ΔE=1/2×F/A×FL/Y

=1/2×10/(10^−4)×(10×0.1)/10^11

ΔE = 5×10^−7J. Ans.

**Que-28: A wire suspended vertically from one end of its end is stretched by attaching a weight of 200 N to the lower end. The weight attached stretches the wire by 1 mm. What is the elastic energy stored in the wire.**

**Solution-** Given F = 200N,

x = 1mm = 10^−3 m

Elastic energy = 1/2 X F X x

∴E = 1/2 X 200 X 1 X 10^−3

= 0.1 J Ans.

—: end of Elasticity Numerical on Bulk Modulus Rigidity and Elastic Potential Energy Class 11 ISC Nootan Solutions Ch-13 Kumar and Mittal Physics :—

Return to : **– Nootan Solutions for ISC Physics Class-11 Nageen Prakashan**

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