# Newton Laws of Motion Numerical on Apparent Weight Class-11 Nootan ISC Physics Solutions

Newton Laws of Motion Numerical on Apparent Weight Class-11 Nootan ISC Physics Solutions Ch-6. 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.

## Newton Laws of Motion Numerical on Apparent Weight

Class-11 Nootan ISC Kumar and Mittal Physics of Nageen Prakashan Solutions

 Board ISC Class 11 Subject Physics Writer Kumar and Mittal Publication Nageen Prakashan Chapter-6 Dimensional Analysis Topics Numerical on Apparent Weight Academic Session 2024-2025

### Numerical on Apparent Weight

Ch-6 Newton Laws of Motion Class-11 ISC Nootan Solutions Kumar and Mittal Physics of Nageen Prakashan

##### Question-11: The strings of a parachute can bear a maximum tension of 72 kg-weight. By what minimum acceleration can a person of 96 kg descend by means of this parachute?

Answer- 96g – 72g = 96a

=>24g = 96a

=>a = 24 / 96 g

=> g/4 = 2.45 m/s²

##### Question-12: A 70 kg person in sea is being lifted by a helicopter with the help of a rope which can bear a maximum tension of 100 kg-weight. With what maximum acceleration the helicopter should rise so that the rope may not break?

Answer- In case of going up T>mg

∴ T -mg = ma

100g – 70g = 70a

30g = 70a

a = 4.2 m/s

##### Question-13: A lift of mass 400 kg is hung by a wire. Calculate the tension in the wire when the lift is (a) at rest, (b) moving upward with a constant velocity of 1.0 m/s, (c) moving upward with an acceleration of 2.0 m/s² and (d) moving downward with an acceleration of 2.0 m/s².

Answer- (a) At rest T -400g = 0

=>T = mg = 400 kg x 9.8 = 3920 N.

(b)again T -400g = 400 x 0 {a = 0}

=> T = 400 x 9.8  = 3920 N.

(c) T – mg = ma

=>T = m (g + a) = 400 kg × (9.8 +2.0) m/s² = 4720 N.

(d) mg – T = ma

=>T= m (g-a)= 400 kg × (9.8-2.0) m/s² = 3120 N

##### Question-14: A body of mass 10 kg is hung by a spring-balance in a lift. What would be the reading of the balance when (i) the lift is ascending with an acceleration of 2 m/s², (ii) descending with the same acceleration, (iii) descending with a constant velocity of 2 m/s? (g = 10 m/s²)

Answer-  (i) W’ – mg = ma

=>W’ = m(g +a) = 10 x 12 = 12 kg-wt

(ii) W’ = m(g + a)

=>W’ = 10 x 8 = 8 kg-wt

(iii) Since acceleration = 0

=>W’ = W = 10 kg-wt

##### Question-15: A lift is going upwards with an acceleration of 4.9 m/s². What will be the apparent weight of a 60 kg man in the lift? What when the lift rises up with a uniform velocity of 4.9 m/s? If the rope of the lift is broken, then?

Answer- W’/W = g + a / g

where W’ = apparent weight

W = real weight

a = acceleration

for going upward  (a = +ve)

∴ W’ / W = 9.8 + 4.9 / 9.8 = 1.5

=>W’ = W x 1.5 = 60 x 1.5 = 90 kg-wt

for uniform velocity a = 0

=> W’ / 60 = g/ g => W’ = 60 kg-wt

for lift falling freely  a =(-9.8)

=> W’ / W = 9.8 – 9.8 / 9.8

=>W’ = 0.

—:  end of Newton Laws of Motion Numerical on Apparent Weight Class-11 ISC Nootan Ch-6 Kumar and Mittal Physics of Nageen Prakashan Solutions. :—

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