ncert solutions physics class 11th
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velocity of a freely falling body is v=
And
-1
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm
velocity of a freely falling body is v=
And
-1
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm
velocity of a freely falling body is v=
And
-1
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm
Explanation- velocity of a freely falling body is v=
And
-1
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm
Explanation- velocity of a freely falling body is v=
And
-1
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6 nm (d). 1.2 x 10 nm
(d)

We can write
OS + PS > OP …….(i)
OS > PS – OP …….(ii)
+ ….(iii)
The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus of both sides as:
> ….(iv)
If the two vectors
6.30 Consider the decay of a free neutron at rest : n p + e .Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the -decay of a neutron or a nucleus (Fig. 6.19).

6.30 Consider the decay of a free neutron at rest : n p + e .Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the -decay of a neutron or a nucleus (Fig. 6.19).

6.30 The decay process of free neutron at rest is given as:
n
From Einstein's mass-energy relation, we have the energy of electron as? where,
? m = Mass defect = Mass of neutron – ( Mass of proton + mass of electron)
c = speed of light
? m and c are constant. Hence, the given two-body
6.29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centers of the balls.

6.29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centers of the balls.

6.29 The potential energy of two masses in a system is inversely proportional to the distance between them. The potential energy of the system of two balls will decrease as they get closer to each other. When the balls touch each other, the potential energy becomes zero, I.e. at r = 2R. The potentia
6.28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
6.28 A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
6.28 Mass, m = 200 kg
Speed, v = 36 km/h = 10 m/s
Mass of the boy, M = 20 kg
Initial momentum = (M + m)v = (20 + 200) x 10 kg-m/s = 2200 kg-m/s
If v' is the final velocity of the trolley, then
The final momentum = (M+m) x v' – M x 4= 220v'-80
According to the law of conservation of energy,
Initial momentu
6.27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?
6.27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?
6.27 Mass of the bolt, m = 0.3 kg, Height of the elevator, h = 3 m
Since the bolt did not rebound, the entire potential energy got converted into heat.
The potential energy of the bolt = mgh = 0.3 x 9.8 x 3 J = 8.82 J
The heat produced will remain same even if the lift is stationary, since g = constant
6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest.
Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest.
Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

6.26 Mass of the block = 1 kg
Spring constant = 100 N/m
Displacement of the block, x = 10 cm = 0.1 m
At equilibrium, normal reaction, R =
Frictional force, F = =
Net force acting on the block down on the incline = - F
= -
=mg ( )
At equilibrium,
6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given = 30 , = 60 , and h = 10 m, what are the speeds and times taken by the two stones ?

6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given = 30 , = 60 , and h = 10 m, what are the speeds and times taken by the two stones ?

6.25 From the law of conservation of energy,
the potential energy at the top = Kinetic energy at the bottom
mgh = (1/2)m ….(1)
and
mgh = (1/2)m ….(2)
= , Both the stones will reach with the same speed
For stone 1, the force acting on the stone 1 is given by , &n
6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
6.24 Mass of the bullet, = 0.012 kg
Initial speed of the bullet, u = 70 m/s
Mass of the wooden block , = 0.4 kg
Initial speed of the wooden block = 0
Let’s assume, final speed of the bullet = v
Applying the law of conservation of momentum
Hence v = ( = 2.04 m/s
Let h be the
6.23 A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.
6.23 A family uses 8 kW of power.
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a typical house.
6.23 Power used by the family = 8 kW = 8000 W
(a) Solar energy received = 200 W/
Percentage conversion of Solar energy to Electrical energy = 20%
If the area required is A then 0.2
A = 200 . The comparable roof size is 14.14 X 14.14 m