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physics ncert exemplar solutions class 12th chapter two

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2 months ago

physics ncert exemplar solutions class 12th chapter two Class 11th

Let $η_{1}$ is the efficiency of an engine at T₁ = $447 ° C$ and T₂ = $147 ° C$ while $η_{2}$ is the efficiency at T₁ = $947 ° C$ and T₂ $47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be

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Payal Gupta

Contributor-Level 10

As we know that

$η = 1 - \frac{T_{2}}{T_{1}}, s o$

$\frac{n_{1}}{n_{2}} = \frac{0.42}{0.74} \approx 0.56$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Class 11th

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt[]{2}$ time the speed of sound, then the value of n will be

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Payal Gupta

Contributor-Level 10

As we know that

$v = \sqrt[]{\frac{γ R T}{M}} \Rightarrow$ Speed of sound in a gas, and

$v_{r m s} = \sqrt[]{\frac{3 R T}{M}} \Rightarrow$ Rms speed of gas molecules, so

$v_{r m s} = \sqrt[]{2} v \Rightarrow \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{2} \sqrt[]{\frac{γ R T}{M}} \Rightarrow γ = \frac{3}{2}$

According formula of specific heat ratio of mixture, we can write

$\Rightarrow$ 5n = 14 n = 2.8

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Oscillations

The length of a seconds pendulum at a height h = 2R from earth surface will be:

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $π^{2} m s^{- 2}$ )

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Payal Gupta

Contributor-Level 10

As we know that

$g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}, s o$

$T = 2 π \sqrt[]{\frac{l}{g_{P}}} \Rightarrow l = \frac{g_{p} T^{2}}{4 π^{2}} = \frac{1}{9} m$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Class 12th

A current of 15 mA flow in the circuit as shown in figure. The value of potential difference between the pointes A and B will be

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Payal Gupta

Contributor-Level 10

$R_{A B} = 5 + (10 / / 5) + 10 = 15 + \frac{10 * 5}{10 + 5} = 15 + \frac{10}{3} = \frac{55}{3} k Ω$

$V_{A B} = R_{A B} l = (\frac{55}{3} * 1 0^{3}) * (15 * 1 0^{- 3}) = 275 V$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Units and Measurement

Velocity $(υ)$ and acceleration (a) in two system of units 1 and 2 are related as $υ_{2} = \frac{n}{m^{2}} υ_{1} a n d a_{2} = \frac{a_{1}}{m n}$ respectively. Here m and n are constants. The relations for distance and time in two system respectively are:

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alok kumar singh

Contributor-Level 10

Let, L₁ & L₂ one distance for system (1) & (2) respectively

T₁ & T₂ are time for system (1) & (2) respectively

Given : v2 = $\frac{n}{m^{2}} v_{1}$

$\Rightarrow (\frac{L_{2}}{T_{2}}) = \frac{n}{m_{2}} (\frac{4}{T_{1}})$ $[? v_{2} = \frac{L_{2}}{T_{2}}, v_{1} = \frac{L_{1}}{T_{1}}]$

$\Rightarrow \frac{L_{2}}{L_{1}} = (\frac{n}{m_{2}}) [\frac{T_{2}}{T_{1}}] - - (i)$

And a2 = $\frac{a_{1}}{m n} \Rightarrow \frac{v_{2}}{T_{2}} = \frac{v_{1}}{T_{1} m n}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \frac{v_{1}}{v_{2}} (\frac{1}{m n})$ $[? \frac{v_{1}}{v_{2}} = \frac{m^{2}}{n}]$

$\Rightarrow \frac{n^{2}}{m} T_{1} = T_{2}$

from (1)

$\frac{L^{2}}{L_{1}} = \frac{n}{m^{2}} [\frac{n^{2}}{m}] [? \frac{T_{2}}{T_{1}} = \frac{n^{2}}{m}]$

$\Rightarrow \frac{n^{3}}{m^{3}} L_{1} = L_{2}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Nuclei

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ is 0.42 V. If the threshold frequency is x * 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 * 10⁸ m/s, Planck's constant = 6.63 * 10^-34 Js)

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Vishal Baghel

Contributor-Level 10

$e V_{0} + h υ_{0} = \frac{h C}{λ}$

$\Rightarrow 1.6 * 1 0^{- 19} * 0.42 + 6.63 * 1 0^{- 34} υ_{0} = \frac{6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{66 3^{0} * 1 0^{- 10}}$

$\Rightarrow 6.63 * 1 0^{- 34} υ_{0} = 1 0^{- 19} (3 - 1.6 * 0.42)$

$6.63 * 1 0^{- 34} υ_{0} = 2.328 * 1 0^{- 19}$

$υ_{0} = 35.11 * 1 0^{13} = 35$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Laws of Motion

A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1cm on main scale and 50 Vernier scale divisions are equal to 49 main scale division, then least count of the travelling microscope is __________* 10^-6 m.

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Vishal Baghel

Contributor-Level 10

Let 'x' be the value of one division on main scale

$x = \frac{1}{40} c m$

Let 'y' be the value of one division on vernier

Now given

50y = 49 x

$y = \frac{49 x}{50}$

$\begin{array}{l} = 5 * 1 0^{- 4} c m \\ = 5 * 1 0^{- 6} m \\ 5 \end{array}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Ray Optics and Optical Instruments

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt[]{7} m .$ The refractive index of water is $\frac{4}{3} .$ The area of the surface of water through which light from the bulb can emerge out is xπm². The value of x is_________.

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Vishal Baghel

Contributor-Level 10

by snell's law

$\frac{4}{3} s i n θ_{C} = μ_{1} s i n 90 °$

$s i n θ_{C} = \frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + {(\sqrt[]{7})}^{2}}}$

$\frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + 7}}$

Squaring both side

$\frac{9}{16} = \frac{r^{2}}{r^{2} + 7}$

$A = r^{2} = π * 9 = 9 π$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Semiconductor Electronics

Two 10cm log, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10^-5N, then separation between the wires is__________cm.

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Vishal Baghel

Contributor-Level 10

$F = \frac{μ_{0} * 2}{4 π} \frac{i_{1} i_{2}}{d} l$

$1 0^{- 5} = \frac{1 0^{- 7} * 2 * 5 * 5}{d} = \frac{10}{100}$

$d = 5 * 1 0^{- 2} m$

d = 5 cm

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2 months ago

physics ncert exemplar solutions class 12th chapter two System of Particles and Rotational Motion

A set of 20 tuning forks is arranged in series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is ___________Hz.

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Vishal Baghel

Contributor-Level 10

A/c to question

f + 76 = 2f

f = 76

last frequency

= f + 76 = 2 * 76

= 152 H₂

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