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ncert solutions physics class 11th

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ncert solutions physics class 11th Physics Gravitation

8.2 Choose the correct alternative:

(a) Acceleration due to gravity increases/decreases with increasing altitude

(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density)

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body

(d) The formula –G Mm(1/r₂ – 1/r₁) is more/less accurate than the formula mg(r₂ – r₁) for the difference of potential energy between two points r₂ and r₁ distance away from the centre of the earth

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Vishal Baghel

Contributor-Level 10

(a) Decreases - Acceleration due to gravity at depth h is given by $g_{h}$ = (1 – $\frac{2 h}{R_{e}}$ )g, where $R_{e} = R a d i u s o f t h e e a r t h,$ g = acceleration due to gravity on the surface of the Earth. From this equation, it is clear that acceleration due to gravity decreases with increase in height

(b) Decreases – Acceleration due to gravity at depth d is given by $g_{d}$ = (1- $\frac{d}{R_{e}}$ )g. So the acceleration due to gravity decreases with increase in depth.

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(b) Decreases – Acceleration due to gravity at depth d is given by $g_{d}$ = (1- $\frac{d}{R_{e}}$ )g. So the acceleration due to gravity decreases with increase in depth.

(c) Mass of the body – Acceleration due to gravity of body mass m is given by the relation g = $\frac{G M}{R^{2}}$ , where G = Universal gravitation constant, M = mass of the Earth and R = radius of the Earth. Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

(d) More – Gravitational potential energy of two points $r_{1}$ and $r_{2}$ distance away from the centre of the Earth is respectively given by:

V( $r_{1}$ ) = $-$ - $\frac{G m M}{r_{1}}$ and V( $r_{2}$ ) = $-$ - $\frac{G m M}{r_{2}}$

Difference in potential energy, V = V( $r_{2}$ ) $-$ V( $r_{1}$ ) = $- G m M (\frac{1}{r_{2}}$ $- \frac{1}{r_{1}}$ )

Hence this formula is more accurate than mg( $r_{2}$ - $r_{1}$ )

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ncert solutions physics class 11th Motion in a Plane

velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $λ = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$λ = h$ -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

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alok kumar singh

Contributor-Level 10

velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $? = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$? = h$ -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

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4 months ago

ncert solutions physics class 11th Motion in a Plane

Explanation- velocity of a freely falling body is v= $\sqrt[]{2 g h}$

And $λ = \frac{h}{m v} = \frac{h}{m \sqrt[]{2 g h}}$

$λ = h$ -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

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alok kumar singh

Contributor-Level 10

(d)

We can write

OS + PS > OP …….(i)

OS > PS – OP …….(ii)

$|\overset{?}{a} - \overset{?}{b}|$ $>$ $|\overset{?}{a}|$ + $|\overset{?}{b}|$ ….(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:

$||\overset{?}{a} - \overset{?}{b}||$ > $||\overset{?}{a}| - |\overset{?}{b}||$ ….(iv)

If the two vectors $\overset{?}{a}$ and $\overset{?}{b}$ act along a straight line but in the opposite direction, then we can write $|\overset{?}{a} - \overset{?}{b}|$ = $|\overset{?}{a}|$ - $|\overset{?}{b}| \dots . . (v)$

Combining (iv) and (v), we get

$|\overset{?}{a} - \overset{?}{b}|$ $\geq$ $|\overset{?}{a}|$ - $|\overset{?}{b}|$

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ncert solutions physics 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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ncert solutions physics 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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ncert solutions physics 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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ncert solutions physics 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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ncert solutions physics 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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ncert solutions physics 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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ncert solutions physics 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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