Fluid Pressure Numerical ISC Class 11 Physics Nootan Solutions Ch-14. 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.

## Fluid Pressure Numerical ISC Class 11 Physics Nootan Solutions Ch-14

Board | ISC |

Class | 11 |

Subject | Physics |

Writer | Kumar and Mittal |

Publication | Nageen Prakashan |

Chapter-14 | Fluid Pressure |

Topics | Numericals on Elasticity, Stress, Strain and Young’s Modulus |

Academic Session | 2024-2025 |

## Numericals on **Elasticity, Stress, Strain and Young’s Modulus of Fluid**

Ch-14 Fluid Pressure ISC Class 11 Physics Nootan Solutions Ch-14 of Kumar and Mittal Physics , Nageen Prakashan

**Que-1: The air pressure is equal to 75cm column of mercury. Express it in N/m^2. ****Given g = 10 m/s^2, density of mercury = 13.6 x 10^3 kg/m^3. **

**Solution- **P = hpg

= 75 x 10^-2 x 13.6 x 10^3 x 10

= 75 x 13.6 x 100

= 1020 x 100

**= 1.02 x 10^5 N/m^2 Ans.**

**Que-2: Calculate the pressure of sea-water at a depth of 200m, when the average density of sea-water is 1.032 x 10^3 kg/m^3. (g = 9.8m/s^2)**

**Solution- **Absolute pressure, P = P_{0 }+ hρg

= (1.013 x 10^{5}) + (200 x 1060) x (9.8)

= (1.013 × 10^{5}) + (20.776 × 10^{5})

= 21.789 × 10^{5
= 2.2 x 10^6 N/m^2 Ans.}

**Que-3: The density of the atmosphere at sea level is 1.29 kg/m^3. Assume that it does not change with altitude, then how high would the atmosphere extend? g = 9.8 m/s^2.**

**Solution- **d = hpg

d = 1.29 × 10 m/s^2

h = 1.29 × 1000 ×10

h = 1.29×10×1000

**h = 7989m Ans.**

**Que-4: If the water pressure gauge shows the pressure at ground floor to be 2.7 x 10^5 N/m^2 then how high would water rise in the pipes of building.**

**Solution- **370×10^{3} = ρgh + 10^{5}

^{pgh = (3.7-1) x 10^5
h = (2.7 x 10^5)/(9.8 x 10^3)
h = 27.6 m Ans.}

**Que-5: The cross-section of two pistons in a hydraulic lifts are 2 cm^2 and 150 cm^2 respectively. Calculate the minimum force required to support a weight of 2000 kg-wt on the broader face of the lift.**

**Solution- **Force = Weight / Area

The area of the larger piston is 150 cm2, and the weight of the object is 2000 kg-wt.

F = 2000 kg-wt / 150 cm^2

F = 13.3 kN

F = 13300 N

F = 13300 / 5

F = 2660 N

**F = 261.3 N Ans.**

**Que-6: A boat having a length of 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm, when a man gets on it. What is the mass of the man.**

Solution- Volume of water displaced = (3 × 2 × 0.01) m3 = 0.06 m3.

∴ Mass of man = Volume of water displaced × Density of water

= (0.06 × 1000) kg = 60 kg Ans.

—: end of Fluid Pressure Numerical ISC Class 11 Nootan Ch-14 Solutions of Kumar and Mittal Physics :—

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

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