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Page No 112: - Chapter 5 Laws Of Motion class 11 ncert solutions Physics - SaraNextGen [2024]


Question 5.26:

A stone of mass tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

 

Lowest Point

Highest Point

(a)

mg – T1

mg + T2

(b)

mg + T1

mg – T2

(c)

mg + T1 –https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_63da9925.gif /R

mg – T2 + https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_63da9925.gif /R

(d)

mg – T1 – https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_63da9925.gif /R

mg + T2 + https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_63da9925.gif /R

 Tand v1 denote the tension and speed at the lowest point. Tand v2 denote corresponding values at the highest point.

Answer:

(a)The free body diagram of the stone at the lowest point is shown in the following figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_2c79a9e9.jpg

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_598a2528.gif  … (i)

Where, v1 = Velocity at the lowest point

The free body diagram of the stone at the highest point is shown in the following figure.

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m65035cc1.jpg

Using Newton’s second law of motion, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4455/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_4ac82e90.gif  … (ii)

Where, v= Velocity at the highest point

It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively (T – mg) and (T + mg).

Question 5.27:

A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

Answer:

(a) Mass of the helicopter, mh = 1000 kg

Mass of the crew and passengers, mp = 300 kg

Total mass of the system, m = 1300 kg

Acceleration of the helicopter, a = 15 m/s2

Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:

R – mpg = ma

mp(g + a)

= 300 (10 + 15) = 300 × 25

= 7500 N

Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.

(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4456/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m48ddaf28.gif

m(g + a)

= 1300 (10 + 15) = 1300 × 25

= 32500 N

The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.

Question 5.28:

A stream of water flowing horizontally with a speed of 15 m s–1 gushes out of a tube of cross-sectional area 10–2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Answer:

Speed of the water stream, v = 15 m/s

Cross-sectional area of the tube, A = 10–2 m2

Volume of water coming out from the pipe per second,

V = Av = 15 × 10–2 m3/s

Density of water, ρ = 103 kg/m3

Mass of water flowing out through the pipe per second = ρ × V = 150 kg/s

The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:

F = Rate of change of momentum https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4457/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m43e8cbce.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4457/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_29c91cd6.gif

Question 5.29:

Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) the force on the 7th coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7th coin by the eighth coin,

(c) the reaction of the 6th coin on the 7th coin.

Answer:

(a) Force on the seventh coin is exerted by the weight of the three coins on its top.

Weight of one coin = mg

Weight of three coins = 3mg

Hence, the force exerted on the 7th coin by the three coins on its top is 3mg. This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin = mg

Weight of the ninth coin = mg

Weight of the tenth coin = mg

Total weight of these three coins = 3mg

Hence, the force exerted on the 7th coin by the eighth coin is 3mg. This force acts vertically downward.

(c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th, and 10th) on its top.

Therefore, the total downward force experienced by the 6th coin is 4mg.

As per Newton’s third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the 7th coin is of magnitude 4mg. This force acts in the upward direction.

Question 5.30:

An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

Answer:

Speed of the aircraft, v = 720 km/h https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4459/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m478467ef.gif

Acceleration due to gravity, g = 10 m/s2

Angle of banking, θ = 15°

For radius r, of the loop, we have the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4459/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_62e2f7bc.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4459/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_m77bec2b2.gif

= 14925.37 m

= 14.92 km

Question 5.31:

A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Answer:

Radius of the circular track, r = 30 m

Speed of the train, v = 54 km/h = 15 m/s

Mass of the train, m = 106 kg

The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail

The angle of banking θ, is related to the radius (r) and speed (v) by the relation:

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4460/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_57eee079.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/4460/NS_14-10-08_Ravinder_11_Physics_5_40_NRJ_html_6625fc95.gif

Therefore, the angle of banking is about 36.87°.

Question 5.32:

A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

https://img-nm.mnimgs.com/img/study_content/curr/1/11/12/183/8087/Chapter%205_html_m2a2c7eb2.jpg

Answer:

750 N and 250 N in the respective cases; Method (b)

Mass of the block, m = 25 kg

Mass of the man, M = 50 kg

Acceleration due to gravity, g = 10 m/s2

Force applied on the block, F = 25 × 10 = 250 N

Weight of the man, W = 50 × 10 = 500 N

Case (a): When the man lifts the block directly

In this case, the man applies a force in the upward direction. This increases his apparent weight.

∴Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley

In this case, the man applies a force in the downward direction. This decreases his apparent weight.

∴Action on the floor by the man = 500 – 250 = 250 N

If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Also Read : Page-No-113:-Chapter-5-Laws-Of-Motion-class-11-ncert-solutions-Physics

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