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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

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2 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven System of Particles and Rotational Motion

A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is $\sqrt[]{x (h^{2} - g L)} .$ The value of x is

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Payal Gupta

Contributor-Level 10

Conservation of Energy b/w (1) & (2)

$\frac{1}{2} m v^{2} + m g l = \frac{1}{2} m u^{2}$

$v = \sqrt[]{u^{2} - 2 g l}$

$\vec{Δ v} = v_{\hat{j}} - u_{\hat{i}}$

$| Δ \vec{v} | = \sqrt[]{{(u)}^{2} + {(\sqrt[]{u^{2} - 2 g l})}^{2}}$

$= \sqrt[]{u^{2} + u^{2} - 2 g l}$

$= \sqrt[]{2} (u^{2} - g l)$

x = 2

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics NCERT Exemplar Solutions Class 12th Chapter Five

One end of a massless spring of spring constant k and natural length $l_{0}$ is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $ω$ about an axis passing through fixed end, then the elongation of the spring will be

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Payal Gupta

Contributor-Level 10

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Motion in a Straight Line

When a ball is dropped into a lake form a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is

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Payal Gupta

Contributor-Level 10

Total time (T) = 4 sec

Given u = 0 m/sec

a = g = 9.8 m/sec²

h = 4.9 m, t =?

$\Rightarrow h = u t + \frac{1}{2} a t^{2}$

$4.9 = 0 . t + \frac{1}{2} * 9.8 t^{2}$

t = 1 sec

'v' be the velocity with which ball hits the water v = u + at

= 0 + 9.8 * 1 = 9.8 m/sec

Time taken to reach the bottom of the lake from surface of the lake

= 4 – 1 = 3 sec

v = 9.8 m/sec

$H = u t + \frac{1}{2} g t^{2} = 9.8 * 3 + \frac{1}{2} * 9.8 * 9$

29.4 + 4.9 * 9 = 29.4 + 44.1

H = 73.5 m

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2 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The distance of the Sun from earth is 1.5 * 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be

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Payal Gupta

Contributor-Level 10

1° → 60'

⇒ 60' → 10°

$60' \to {\frac{π}{180}}^{C}$

$θ = \frac{π * 2 * 1 0^{3}}{180 * 3600}$

L = 1.5 * 10¹¹ m

$D = L θ = 1.5 * 1 0^{11} * \frac{π}{180 * 3600} * 2 * 1 0^{3} = 1.45 * 1 0^{9} m$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The distance of the Sun from earth is 1.5 × 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Ray Optics and Optical Instruments

The distance of the Sun from earth is 1.5 × 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be:

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 11th

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

(a) The rate of cooling is constant till milk attains the temperature of the surrounding

(b) The temperature of milk falls off exponentially with time

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.

According to newton's law of cooling temperature of the milk falls of exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to surroundings. So option b, c, d satisfy.

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 11th

Refer to the plot of temperature versus time (Fig) showing the changes in the state of ice on heating (not to scale) Which of the following is correct?

(a) The region AB represents ice and water in thermal equilibrium

(b) At B water starts boiling

(c) At C all the water gets converted into steam

(d) C to D represents water and steam in equilibrium at boiling point

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) During the process AB temperature of the system is 0⁰C . hence it represents phase change that is transformation of ice into water while temperature remains 0⁰C.

BC represents rise in temperature of water from 0⁰C to 100⁰C. now water starts converting into steam which is represented by CD.

Which is only represents by a, d options

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 11th

'Gulab Jamuns' (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice big (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following:

(a) Both size gulab jamuns will get heated in the same time

(b) Smaller gulab jamuns are heated before bigger ones

(c) Smaller pizzas are heated before bigger ones

(d) Bigger pizzas are heated before smaller ones

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Smaller gulab jamun having least surface area hence they will be heated first. As in case of smaller gulab jamun heat radiates will be less.

Similarly smaller pizzas are heated before bigger ones because of smaller surface area.

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4 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 11th

Mark the correct options:

(a) A system X is in thermal equilibrium with Y but not with Z. System Y and Z may be in thermal equilibrium with each other

(b) A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other

(c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other

(d) A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z may be in thermal equilibrium with each other

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) T_x=T_y when x and y are in thermal equilibrium.

T_{x
$\neq$}T_Zwhen x and z are not in thermal equilibrium.

T_{y
$\neq$}T_Z

Hence y and z are not In thermal equilibrium.

Given T_{x
$\neq$}T_y

And T_{x
$\neq$}T_Z

We cannot say about equilibrium of y and z.

They may or may not in thermal equilibrium

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