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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 Dual Nature of Radiation and Matter

An α particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be

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Payal Gupta

Contributor-Level 10

$k_{α} = 2 e V \Rightarrow P_{α} = \sqrt[]{2 K_{α} m_{α}}$

k_P = eV P_P = $\sqrt[]{2 K_{P} m_{P}}$

$∴ \frac{P_{α}}{P_{P}} = \sqrt[]{\frac{K_{α} m_{α}}{K_{P} m_{P}}} = \sqrt[]{\frac{2 * (4 m_{P})}{m_{P}}} \begin{matrix} \sqrt[]{8} : 1 & 2 \sqrt[]{2} : 1 \end{matrix}$

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2 months ago

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

An unpolarised light beam of intensity 2l₀ is passed through a polaroid P and then through another polaroid Q which is oriented is such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

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Payal Gupta

Contributor-Level 10

$I = I_{0} c o s^{2} 30 °$

$= I_{0} {(\frac{\sqrt[]{3}}{2})}^{2} = \frac{3}{4} I_{0}$

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2 months ago

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

A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by $\frac{π}{2} .$ If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

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Payal Gupta

Contributor-Level 10

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$

$5 = \frac{100}{R_{y}} \Rightarrow R_{y} = \frac{100}{5} = 20 Ω$

When x and y both are connected in series :-

v = 100sin $ω t$

tan = 1 = 45°

R₀ = $\sqrt[]{R_{x}^{2} + R_{y}^{2}} = 20 \sqrt[]{2} Ω$

$∴ \overset{o}{l_{0}} = \frac{v_{0}}{R_{0}} = \frac{100}{20 \sqrt[]{2}} = \frac{5}{\sqrt[]{2}} A .$

$∴ l_{m s} = \frac{l_{0}}{\sqrt[]{2}}$

$= \frac{5}{\sqrt[]{2} \cdot \sqrt[]{2}}$

$= \frac{5}{2} A = \frac{5}{2} A$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Motion in Straight Line

A juggler throws ball vertically upwards with same initial velocity in air. When the first ball reaches its highest positions he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

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Payal Gupta

Contributor-Level 10

Time to reach at max height t = $\frac{u}{g} n o .$ of balls thrown in 1 sec = n.

So, time taken by each ball to reach maxm height, = $\frac{1}{n} s e c$

i.e. $\frac{u}{g} = \frac{1}{n} \Rightarrow u = \frac{g}{n}$ So, hmax = $\frac{u^{2}}{2 g}$

= $\frac{g^{2}}{2 g n^{2}}$

$= \frac{g}{2 n^{2}}$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Moving Charges and Magnetism

A wire X of length 50cm carrying a current of 2A is placed parallel to a long wire Y of length 5m. The wire Y current of 3A. The distance between two wires is 5cm and currents flow in the same direction. The force acting on the wire Y is

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Payal Gupta

Contributor-Level 10

F_x-y = $\frac{μ_{0} i_{1} i_{2}}{2 π r} * (. 5)$

$= \frac{2 * 1 0^{- 7} * 6 * 0.5}{0.05}$

$= \frac{6}{5} * 1 0^{- 5} N = 1.2 * 1 0^{- 5}$

Force on Y = 1.2 * 105N towards 'x'

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Atoms

Hydrogen atom from exited state comes to the ground state by emitting a photon of wavelength $λ .$ The value of principal quantum number 'n' of the excited state will be

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Payal Gupta

Contributor-Level 10

According to definition of wave number, we can write

$\frac{1}{λ} = R (\frac{1}{1^{2}} - \frac{1}{n^{2}}) \Rightarrow 1 - \frac{1}{n^{2}} = \frac{1}{λ R}$

$\Rightarrow \frac{1}{n^{2}} = 1 - \frac{1}{λ R} = \frac{λ R - 1}{λ R} \Rightarrow n = \sqrt[]{\frac{λ R}{λ R - 1}}$

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2 months ago

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

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are 1%, 2% and 3% respectively. The maximum percentage error in the detection of the dissipated heat will be

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Payal Gupta

Contributor-Level 10

$H = l^{2} R t$

$% e r r o r i n H = \frac{Δ H}{H} * 100 = 2 (\frac{Δ l}{l}) * 100 + (\frac{Δ R}{R}) * 100 + (\frac{Δ t}{t}) * 100 = 2 * 2 + 1 + 3 = 8 %$

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2 months ago

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

A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

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Payal Gupta

Contributor-Level 10

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Work, Energy and Power

A bag of sand of mass 9.8kg is suspended by a rope. A bullet of 200g travelling with speed 10 ms^-1 gets embedded in it, then loss of kinetic energy will be

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Payal Gupta

Contributor-Level 10

$v = \frac{m v_{0}}{M + m} \Rightarrow K_{f} = \frac{1}{2} (M + m) v^{2} = \frac{m v_{0}^{2}}{2 (M + m)}$

$K_{i} = \frac{m v_{0}^{2}}{2}$

Loss in Kinetic energy = $Δ K = K_{i} - K_{f}$

$Δ K = \frac{1}{2} m v_{0}^{2} (\frac{M}{M + m}) = \frac{1}{2} * 0.2 * 100 = [\frac{9.8}{10}] = 9.8 J$

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2 months ago

physics ncert exemplar solutions class 12th chapter two Physics Gravitation

An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, R = 6400km. The percentage decrease in the weight of the object will be

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Payal Gupta

Contributor-Level 10

As we know that $g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{\frac{5}{4} R})}^{2} = \frac{16 g}{25}$

$\Rightarrow Δ g = g - g_{P} = \frac{9 g}{25}$

The percentage decrease in the weight of the object = $\frac{Δ g}{g} * 100 = 36 %$

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