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physics ncert solutions class 11th

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physics ncert solutions class 11th Work, Energy and Power

6.30 Consider the decay of a free neutron at rest : n $\to$ 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 $β$ -decay of a neutron or a nucleus (Fig. 6.19).

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Payal Gupta

Contributor-Level 10

6.30 The decay process of free neutron at rest is given as:

n $? p + \overset{?}{e}$

From Einstein's mass-energy relation, we have the energy of electron as? $m c^{2}$ 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 decay is unable to explain the continuous energy distribution in the $?$ -decay of a neutron or a nucleus. The presence of neutrino von the LHS of the decay correctly explain the continuous energy distribution.

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physics ncert solutions class 11th Work, Energy and Power

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.

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Payal Gupta

Contributor-Level 10

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 potential energy curve in (i), (ii), (iii), (iv) and (vi) do not satisfy these conditions. So there is no elastic collision.

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physics ncert solutions class 11th Work, Energy and Power

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?

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Payal Gupta

Contributor-Level 10

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 momentum = final momentum

2200 = 220v'-80

V' = 10.36 m/s

Time required by the boy to travel 10m = 10/4 = 2.5 s

Distance travel by trolley in 2.5 s = 10.36 x 2.5 m = 25.9 m

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physics ncert solutions class 11th Work, Energy and Power

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 ?

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Payal Gupta

Contributor-Level 10

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

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physics ncert solutions class 11th Work, Energy and Power

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.

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Payal Gupta

Contributor-Level 10

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 = $m g \cos ? 37 °$

Frictional force, F = $? R$ = $m g \sin ? 37 °$

Net force acting on the block down on the incline = $m g \sin ? 37 °$ - F

= $m g \sin ? 37 °$ - $? m g \cos ? 37 °$

=mg ( $\sin ? 37 ° - ? \cos ? 37 °$ )

At equilibrium,

Work done = Potential energy of the stretched string

mg ( $\sin ? 37 ° - ? \cos ? 37 °$ ) = (1/2)kx2

1 x 10 x ( $\sin ? 37 ° - ? \cos ? 37 °$ ) = (1/2) x 100 x 0.1

10 x (0.602 – $? 0.798)$ = 0.5 x 100 x 0.1

$? = 0.128$

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physics ncert solutions class 11th Work, Energy and Power

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 $θ_{1}$ = 30 $°$ , $θ_{2}$ = 60 $°$ , and h = 10 m, what are the speeds and times taken by the two stones ?

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Payal Gupta

Contributor-Level 10

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ &

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6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ = v/ $a_{1}$ = v/g $\sin ? ?_{1}$ = 14/(9.81 $\sin ? 30 °)$ = 2.85 s

$t_{2}$ = v/ $a_{2}$ = v/g $\sin ? ?_{2}$ = 14/(9.81 $\sin ? 60 °)$ = 1.65 s

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physics ncert solutions class 11th Work, Energy and Power

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.

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Payal Gupta

Contributor-Level 10

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 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

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2

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6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 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

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2) $*$ 0.012 X $70^{2}$ - (1/2) $* (0.012 + 0.4) * {2.04}^{2}$ J

= 29.4 – 0.857 J

= 28.54 J

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physics ncert solutions class 11th Work, Energy and Power

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.

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Payal Gupta

Contributor-Level 10

6.23 Power used by the family = 8 kW = 8000 W

(a) Solar energy received = 200 W/ $m^{2}$

Percentage conversion of Solar energy to Electrical energy = 20%

If the area required is A then 0.2 $* A * 200 = 8000$

A = 200 $m^{2}$ . The comparable roof size is 14.14 X 14.14 m

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physics ncert solutions class 11th Motion in a Plane

Q.4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 ? m/s and moves in the x-y plane with a constant acceleration of (8.0 ? + 2.0 ?) m s^-2.

(a) At what time is the x- coordinates of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

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Vishal Baghel

Contributor-Level 10

Ans.4.21:

(a) Velocity , $\overset{?}{v}$ = 10.0 ? m/s

Acceleration, $\overset{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = 8.0 ? + 2.0 ?

$\overset{?}{d v}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$ ,

Where, $\overset{?}{u} =$ velocity vector of the particle at t =0

$\overset{?}{v} =$ velocity vector of the particle at time t

But $\overset{?}{v}$ = $\frac{\underset{d r}{?}}{d t}$

$\overset{?}{d r}$ = $\overset{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\overset{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\overset{?}{r} =$ $\overset{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\overset{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\overset{?}{u}$ , we get

$\overset{?}{r} =$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x / 4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$

At t = 2 s,

$\overset{?}{v}$ (t) = 8.0 $*$ 2 ? + 2.0 $* 2$ ? + 10 $*$ ? = 16 ? + 14 ?

The magnitude of $\overset{?}{v}$ (t) is given by

$|\overset{?}{v}|$ = ( 162 + 142)1/2 = 21.26 m/s

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physics ncert solutions class 11th Work, Energy and Power

6.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

(a) How much work does she do against the gravitational force ?

(b) Fat supplies 3.8 * 10⁷J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

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Payal Gupta

Contributor-Level 10

6.22 Mass lifted, m = 10 kg

Height to which the mass lifted, h = 0.5 m

No of repetitions, n = 1000

(a) Work done against gravitational force,

W = nmgh = 1000 $* 10 * 9.81 * 0.5 =$ 49050 J

(b) Mechanical energy supplied by 1 kg fat, with 20% efficiency rate = 0.2 $*$ 3.8 $* 10^{7}$ = 0.76 $* 10^{7}$ J/kg

Fat used by dieter = 49050 / (0.76 $* 10^{7})$ kg = $6.45 * 10^{- 3}$ kg

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