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physics ncert solutions class 11th

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physics ncert solutions class 11th Class 11th

A refrigerator consumes an average 35W power to operate between temperature -10°C to 25°C. If there is no loss of energy then how much average heat per second does it transfer?

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alok kumar singh

Contributor-Level 10

$\frac{Q_{H}}{Q_{L}} = \frac{T_{H}}{T_{L}}$

$\Rightarrow Q_{L} = \frac{T_{L}}{T_{H} - T_{L}} * (Q_{H} - Q_{L})$

$\Rightarrow \frac{d Q_{L}}{d t} = \frac{T_{L}}{T_{H} - T_{L}} . \frac{d (Q_{H} - Q_{L})}{d t}$

$= \frac{273 - 10}{25 - (- 10)} * 35$

$= 263 J s^{- 1}$

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2 months ago

physics ncert solutions class 11th Class 11th

The rms speed of the molecules of Hydrogen, Oxygen and Carbondioxide at the same temperature are $V_{H}, V_{O} a n d V_{C}$ respectively then:

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Vishal Baghel

Contributor-Level 10

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}}$

$V_{r m s} \propto \frac{1}{\sqrt[]{M}}$ at constant T

$V_{H} : V_{O} : V_{C o_{2}} = \frac{1}{\sqrt[]{1}} : \frac{1}{\sqrt[]{16}} : \frac{1}{\sqrt[]{44}}$

$= 1 : \frac{1}{4} : \frac{1}{2 \sqrt[]{11}}$

$V_{H} > V_{O} > V_{C O_{2}}$

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2 months ago

physics ncert solutions class 11th Class 11th

Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by X_p (t) = at + bt² and X_Q (t) = ft – t². At what time, both the buses have same velocity?

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Raj Pandey

Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ - (ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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physics ncert solutions class 11th Class 11th

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Raj Pandey

Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ - (ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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physics ncert solutions class 11th Class 11th

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then $R = 4 \sqrt[]{h_{1} h_{2}}$

Reason R : Product of said heights.

$h_{1} h_{2} = (\frac{u^{2} s i n^{2} θ}{2 g}) . (\frac{u^{2} c o s^{2} θ}{2 g})$

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Raj Pandey

Contributor-Level 9

First angle, 1 = $R = \frac{u^{2} s i n 2 ?}{g}$ $\frac{^{}}{}$

Another angle (2 = 90 - ) for which range will be

Same as that of 1 =

$R^{1} = \frac{u^{2} s i n 2 (90 ? ?)}{g} = \frac{u^{2}}{g} s i n 2 ? = R$

at $?_{1} = ?, ? ? ? ? ? h_{1} = \frac{u^{2} s i n^{2} ?}{2 g}$ $_{}_{} \frac{^{}^{}}{}$

$& ? ? ? ?_{2} = 90 ? ?, h_{2} = \frac{u^{2} s i n^{2} ?_{2}}{2 g} = \frac{u^{2} c o s^{2} ?}{2 g}$

$? h_{1} h_{2} = \frac{1}{4^{2}} {[\frac{u^{2} s i n 2 ?}{g}]}^{2}$

$h_{1} h_{2} = \frac{1}{4^{2}} R^{2}$

$R = 4 \sqrt[]{h_{1} h_{2}}$

So, Both statement is true & Reason is correct explanation for statement 1.

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2 months ago

physics ncert solutions class 11th Class 11th

Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is 1 : 4, then

(A) The r.m.s. velocity of gas molecules in two vessels will be the same.

(B) The ratio of pressure in these vessels will be 1 : 4.

(C) The ratio of pressure will be 1 : 1.

(D) The r.m.s. velocity of gas molecules in two vessels will be 1 : 4.

Choose the correct answer form the options given below:

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alok kumar singh

Contributor-Level 10

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} A s . v_{1} = v_{2}$

$V_{r m s} α \sqrt[]{T}$ $η_{2} = 4 n_{1}$

$T_{1} = T_{2} \Rightarrow V_{r m s - 1} = V_{r m s - 2}$

$P_{1} = \frac{n_{1} R T}{v} P_{z} = \frac{n_{2} R T}{v}$

$\frac{P_{1}}{P_{2}} = \frac{n_{1}}{n_{2}} = \frac{1}{4}$

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2 months ago

physics ncert solutions class 11th Class 11th

The relation between root mean square speed $(υ_{r m s})$ and most probable speed $(υ_{p})$ for the milar mass M of oxygen gas molecule at the temperature of 300K will be:

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Vishal Baghel

Contributor-Level 10

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}}$

& $v_{P} = \sqrt[]{\frac{2 R T}{M}}$

$v_{r m s} = \sqrt[]{\frac{3}{2}} v_{P}$

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physics ncert solutions class 11th Class 11th

The root mean square speed of smoke particles of mass 5 * 10^-17kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 * 10^-23 JK^-1]

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Payal Gupta

Contributor-Level 10

m = 5 * 10^-17kg ; k = 1.38 * 10^-23 Jk^-1

$V_{r m s} = \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{\frac{3 N k T}{N m}} = \sqrt[]{\frac{3 k T}{m}} = \sqrt[]{\frac{3 * 1.38 * 1 0^{- 23} * 300}{5 * 1 0^{- 17}}}$

$= \sqrt[]{18 * 13.8 * 1 0^{- 6}}$

$= 15.7 * 1 0^{- 3} m / s$

$= 15 m m / s$

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2 months ago

physics ncert solutions class 11th Class 11th

For a particle in uniform circular motion, the acceleration $\vec{a}$ at any point P(R, q) on the circular path of radius R is (when q is measured from the positive x-axis and v is uniform speed):

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

$a = \frac{v^{2}}{R}$

$\vec{a} = - \frac{v^{2}}{R} c o s θ \hat{i} - \frac{v^{2}}{R} s i n θ \hat{j}$

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2 months ago

physics ncert solutions class 11th Class 11th

The torque of a force $5 \hat{i} + 3 \hat{j} - 7 \hat{k}$ about the origin is $τ$ . If the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + \hat{k},$ then the value of $τ$ will be

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Payal Gupta

Contributor-Level 10

$\vec{F} = 5 \hat{i} + 3 \hat{j} - 7 \hat{k}$

$\vec{r} = 2 \hat{i} + 2 \hat{j} + \hat{k}$

$\vec{τ} = | \vec{r} * \vec{F} | = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 \\ 5 & 3 & - 7 \end{matrix} | \begin{matrix} = \hat{i} (- k_{1} - 3) - \hat{j} (- 14 - 5) + \hat{k} (6 - 10) & = - 17 \hat{i} + 19 \hat{j} - 4 \hat{k} \end{matrix}$

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