Tags Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

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2 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Class 12th

The electric field in a region is given by $\vec{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \frac{N}{C} .$ The ratio of flux of reported field through the rectangular surface of are 0.2m² (parallel to y – z plane) to the of the surface of are 0.3m² (parallel to x – z plane) is a : b, where a = ……………

[Here $\hat{i}, \hat{j} a n d \hat{k}$ are the unit vectors along x, y and z – axes respectively]

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Vishal Baghel

Contributor-Level 10

$?_{1} = \frac{3}{5} E_{0} (0.2) N m^{2} C^{- 1}, a n d ?_{2} = \frac{4}{5} E_{0} (0.3) N m^{2} C^{- 1}$

$\Rightarrow \frac{?_{1}}{?_{2}} = \frac{3 * 0.2 * 5}{5 * 0.3 * 4} = \frac{1}{2}$

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2 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Work, Energy and Power

A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ……………… m/s. (Take g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

Let v is velocity at the highest position.

$T_{m a x} = 5 T_{m i n} \Rightarrow m g + \frac{m (v^{2} + 4 g l)}{l} = 5 (\frac{m v^{2}}{l} - m g) \Rightarrow 4 . \frac{v^{2}}{l} = 10 g$

$\Rightarrow v = \sqrt[]{\frac{5}{2} g l} = \sqrt[]{\frac{5}{2} * 10 * 1} = 5 m / s$

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2 weeks ago

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

A mono-atomic gas of mass 4.0u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is $\frac{x}{3 R} .$ Value of x is…………….

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Vishal Baghel

Contributor-Level 10

$Δ U + Δ K E = 0$

$\Rightarrow \frac{f}{2} n R Δ T = \frac{1}{2} m v^{2} \Rightarrow Δ T = \frac{m}{n} \frac{v^{2}}{f R} = 4 * 1 0^{- 3} * \frac{3 0^{2}}{3 R} = \frac{3.6}{3 R} K .$

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2 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Physics Oscillations

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as $U = \frac{α}{r^{10}} - \frac{β}{r^{5}} - 3$

Where, α and β are positive constants. The equilibrium between two atoms will be ${(\frac{2 α}{β})}^{\frac{a}{b}},$ where α =…………….

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Vishal Baghel

Contributor-Level 10

For equilibrium,

$\frac{d U}{d r} = 0 \Rightarrow \frac{- 10 α}{r^{11}} + \frac{5 β}{r^{6}} = 0 \Rightarrow r = {(\frac{2 α}{β})}^{\frac{1}{5}}$

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Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Electromagnetic Induction

A coil of inductance 2H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, the energy stored in the coil after 4s is…………… J.

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Vishal Baghel

Contributor-Level 10

$L \frac{d i}{d t} = V \Rightarrow 2 d i = 3 t d t \Rightarrow 2 \int_{0}^{i} d i = 3 \int_{0}^{4} t d t \Rightarrow i = 12 A$

$U = \frac{1}{2} L i^{2} = \frac{1}{2} * 2 * 1 2^{2} = 144 J$

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3 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Class 12th

A point object moves towards a concave mirror with a speed $10 c m s^{- 1}$ along the principal axis. The magnitude of the velocity of object with respect to image when the object is at distance of 15cm from the pole is __ $c m s^{- 1}$ . (Focal length of the mirror kept at rest is 10cm )

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Raj Pandey

Contributor-Level 9

$v = \frac{f u}{u - f} = \frac{(- 10) (- 15)}{- 15 + 10} c m = - 30 c m$

$v_{image} = \frac{d v}{d t} = - {(\frac{v}{u})}^{2} \frac{d u}{d t} = - {(\frac{- 30}{- 15})}^{2} (10) c m s^{- 1} = - 40 c m s^{- 1}$

$V_{object} = \frac{d u}{d t} = + 10 c m s^{- 1}$

Relative velocity $= v_{object} - v_{image}$

$= 10 - (- 40) = 50 c m s^{- 1}$

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3 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Physics Thermodynamics

Two carnot engines operate in series between a heat source at a temperature $T_{1} = 900 K$ and heat sink at temperature $T_{3} = 400 K$ . There is another reservoir at temperature $T_{2}$ , as shown, with $T_{1} > T_{2} > T_{3}$ . The two engines are equally efficient if $T_{2} =$ K

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Raj Pandey

Contributor-Level 9

$η_{1} = 1 - \frac{T_{2}}{T_{1}}, η_{2} = 1 - \frac{T_{3}}{T_{2}}$

as $η_{1} = η_{2} \Rightarrow \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} \Rightarrow T_{2} = {(T_{1} T_{3})}^{\frac{1}{2}} = \sqrt[]{400 * 900} = 600 K$

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3 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Physics Magnetism and Matter

A paramagnetic sample shows a net magnetisation of $12 A / m$ when placed in an external magnetic field of $0.6 T$ at a temperature of $4 K$ . When the same sample is placed in an external magnetic field of $0.2 T$ at temperature of 16 k, the net magnetisation will be $A / m$ .

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Raj Pandey

Contributor-Level 9

$M = χ_{m} H$ and $_{} \frac{}{}$ $χ_{m} = \frac{C}{T} ($ Curie law $)$ ,

$M = \frac{C H}{T}$ or $\frac{M_{1}}{M_{2}} = \frac{C H_{1} / T_{1}}{C H_{2} / T_{2}} = (\frac{H_{1}}{H_{2}}) (\frac{T_{2}}{T_{1}})$ $\frac{_{}}{_{}} \frac{_{}_{}}{_{}_{}} (\frac{_{}}{_{}}) (\frac{_{}}{_{}})$

$M_{2} = M_{1} (\frac{H_{2}}{H_{1}}) (\frac{T_{1}}{T_{2}}) = 12 (A / m) [\frac{0.2 T}{0.6 T}] [\frac{4 K}{16 K}]$

$= 1 A / m$

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3 weeks ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen Physics Thermodynamics

A hot body is being cooled in air according to Newton's law of cooling, the rate of fall of temperature being K times the difference of its temperature with respect to that of surroundings. The time after which the rate of heat loss of the body will be half of the maximum rate of heat loss is $\frac{1}{2 k} {l o g}_{e} ? n$ (The time is to be counted from the instant t = 0 ). The value of n is

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Raj Pandey

Contributor-Level 9

Newton's law of cooling, $\begin{array}{r} - \frac{d T}{d t} = k (T - T_{0}) \Rightarrow - \int \frac{d T}{T - T_{0}} = k \int d t \\ \Rightarrow {[- l n ? |T - T_{0}|]}_{T}^{(T + T_{0}) / 2} = k t (? Heat \propto (T - T_{0})) \\ \Rightarrow - l n ? |\frac{(T - T_{0}) / 2}{T - T_{0}}| = k t \Rightarrow - l n ? |\frac{1}{2}| = k t \Rightarrow t = \frac{l n ? 2}{k} = \frac{l n ? 4}{2 k} ∴ n = 4 \end{array}$

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3 weeks ago

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

A 1kg particle is moving unidirectionally on a horizontal plane under the action of power $P = - 2 v^{3}$ , of power supplying energy source. The velocity (v) - time (t) graph that describes the motion of the particle is shown in figure. (Graphs are drawn schematically and are not to scale) The value of $t_{0}$ in graph is sec. Find 8x