Physics Revision Sheet XII IIT by Sumit Upmanyu Sir

REVISION SHEET “Take Subject : PHYSICS

GOAL

up

ONE

in

your

LIFE. Make that Topic :

One Goal your life-

XI - Syllabus

Think of it, Dream of it, Live on that

Revision Sheet :

To be Solved in

days.

GOAL.

Let

the

Brain,

Muscles,

Nerves, Every part if your body, be full of that Goal, and just leave every other thing alone. This is the only way to SUCCESS, that is the way great people are

produced.” :

- SWAMI VIVEKANAND Q.1

(a) (b) (c) Q.2

(a) (b) (c)

A particle with mass 2 kg moves in one dimension, in the presence of a force that is described by the following potential energy graph. In parts (a)-(c), specify your numerical answers to 2 significant figure accuracy. If the particle is located at x = 0.5 m, what is Fx, the x-component of the force acting on the particle? If the particle is located at x = 2 m, what is Fx, the x-component of the force acting on the particle? If the particle is released from rest at x = 2 m, what will be its speed when it crosses x = 5 m? An initially empty beaker, in the shape of a cylinder with cross sectional area A, is left out in the rain. The raindrops hit the beaker vertically downward with speed v. The rain continues at a constant rate, so the height of the water in the beaker h(t) increases with time t at a rate dh /dt = w, where w is negligible compared to v. The raindrops quickly come to rest inside the beaker, so we can neglect any kinetic energy of the water that has collected in the beaker. Let  denote the density of water (i.e., the mass per unit volume). What is the rate at which the mass of the water in the beaker increases with time? Let ycm(t) denote the height of the center of mass of all the water that has collected in the beaker by time t. What is the rate dy cm dt at which this height increases?   The total momentum of any system of particles Ptot is equal to the total mass Mtot times the velocity v cm of the center of mass. Should we conclude, there fore, that the water in the beaker has a vertical momentum equal to its mass times the value of dy cm dt as described in part (b)? Explain your answer in one or two sentences.

(d)

If the beaker is placed on a scale, while the beaker is still in the rain, the impact of the raindrops on the beaker will cause the reading on the scale to be larger than the weight of the beaker and the water it contains. By how much is the reading on the scale increased by the impact of the raindrops? (Neglect the effect of raindrops that hit the scale directly.) CatalyseR Eduventures (India) Pvt.

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Q.3

(a) (b)

(c) (d) (e)

(f)

(g)

Q.4

(a) (b)

A volume V0 of helium is confined within a chamber. The initial pressure is PA (atmospheric pressure), and the initial temperature is T0. Helium is a monatomic gas, which we will treat as an ideal gas. Each atom of helium has a mass mHe = 4 u, where 1 u = 1.66 × 10–27 kg. What is the number N0 of atoms of helium in the sample. Your answer should be expressed exclusively in terms of the given quantities, but it may also contain k (the Boltzmann constant) or R(the gas constant). What is the total internal energy U0 of the gas? Suppose the gas is allowed to expand to V1 = 1.5 V0, while it is being heated at just the right rate to keep the pressure constant. What is the temperature T1 of the gas at the end of this expansion? During the expansion, how much work W1 is done by the gas? During the expansion, how much heat Q1 has to be added to the gas. The gas is then compressed back to its original volume V0, with the compression taking place at the constant temperature T1. How much work W2 is done on the gas during this compression. The gas is allowed to cool back to temperature T0, so the pressure returns to PA and the initial conditions are restored. A second chamber, also of volume V0, contains neon at pressure PA and temperature T2. Neon is also a monatomic gas, which we will treat as ideal. The mass of each neon atom is mNe = 20 u. The two chambers are connected by a tube (of negligible volume), so that the gases can mix. The entire system is insulated, so that no heat escapes. Once equilibrium is reached, what is the temperature T3 of the mixture of gases? A particle moves with uniform speed of m/s along a path comprising of two semicircles from A to B and then from B to C as shown in figure. Find magnitude of average velocity for the entire journey 3 Find magnitude of displacement of the particle from A, sec 2 after start

Q.5

A pressure cooker is a pan whose lid can be tightly sealed to prevent gas from escaping. Suppose an empty pressure cooker is left on a hot stove, so that the temperature inside rises to T1. A pressure relief value on the lid is then opened, allowing hot air to escape until the pressure inside is reduced to atmospheric. The pot is then sealed, and removed from the stove and allowed to cool. What is the net force due to air pressure on the lid when the contents have cooled to room temperature, T0? Be sure to give the magnitude and direction. You may assume that the lid of the pressure cooker is in the shape of a circle of diameter l, and that the atmospheric pressure is P0. Both temperatures T1 and T0 are measured on an absolute scale, such as the Kelvin scale.

Q.6

A bullet of mass m1 is fired into a pendulum of mass m2 and length L. The speed of the bullet as it enters the mass m2 is V1 (see figure)

(a) (b)

First, assume that the collision is elastic, and that m1 > m1, the bullet will leave with the same relative speed with which it came in V1'  V1 (b)

5.

1 2 1  T0  l P0  T  4  1 

The speed of the bullet after the collision is now V1 + V2 relative to m2. Therefore, the speed of

m1 is V1'  V1  2V2 . (c)

m1V1 Momentum conservation gives the speed of both masses after the collision, V’ = m  m . This kinetic 1 2 energy brings the system to a height given by 2 2 1 m1 V1 1 r2 (m + m2)V = (m1 + m2)gh ; 2 = gL (1–cos max) ( m1  m 2 ) 2 2 1

 V1 = (d)

m1  m 2 m1

2gL (1  cos max ) Notice for  =0, V = 0 as it should! max 1

If max = 90°, then cos max = 0 and V1=

m1  m 2 2gL , which is possible. For example, if L = 1 m m1

m2 and m 102 , then V1  450 m/sec.] 1 3 (T1  mg ) ; (c)

3(T1  mg )l m

7.

(a) T2 = T1– 2mg; (b)

8.

1 + cos

9.

(a)AC, (b)170J, (c) 10J

10.

 vt  v2 2h Mg F lF  (a) F = , (b) l – , (c) v = sin , (d) x = v , (e) ax = l sin  sin  l sin   g cos  k M  

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200 40 40 ˆ 30 ˆ 70 ˆ ˆi , a m1  i j , a m2   j N, aM = 17 17 17 17 17

11.

T=

12. 13. 14. 15. 16.

g/11 MB = 15 kg, T = 15 , aA = 5 m/s2 , aB = 9 m/s2 55 sec (a) 2 ms–2, (b) 2.4 N, 0.3 (c) 0.2 s 5g, 5g/2, 0

17.

3.3 m

19.

(a)

21.

2mv 2 3l

22.

5.0 J, Total loss of mechanical energy E = E1 + E2 = 30, string becomes right after

3gL , (b) mgL [2+ 3 ]

18.

 mM   F = 2smg  2 m  M  

20.

4 m/s, 24.5J, 40 J

1 sec, 2m g 1 ; (c) E = l2g 3 6

23.

(b)

24.

v2 =

( 2eM  3m ) v 0 3 (1  e)mv 0 ; v1= 2M  3m 2M  3m

25.

R=

1 1 g(L + 3x), T1 = gx, Q = gL2 2 4

26.

(a) v0, (b)

27.

4v 0 , (c) x = vo 5

v0 g sin 

28.

=

6k m

29.

=

 12k  48v 2 m t sin  2  m l 49k  

30.

(a) T= 2

31. 32. 33. 34. 35.

m 5k

12 5R 5R g(1  cos 0 ) , (c) 2 , (b) 5 3g 3g

13 mg 7 (i) 200 J/°C (ii) 0.5 °C/sec (iii) 50W steel, 500/9 °C (a) (dgL)/4Y, (b) (dgL)/6Y

2m 2 2gr m1  m 2

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1 sec, 3