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Physics NCERT Exemplar Solutions Class 12th Chapter Three

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Physics Thermodynamics

In an experiment to verigy Newton's law of cooling, a graph is plotted between, the temperature difference $(Δ T)$ of the water and surrounding and time as shown in figure. The initial temperature of water is taken as $80 ° C .$ The value of t₂ as mentioned in the graph will be _______.

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Payal Gupta

Contributor-Level 10

$\frac{d T}{d t} = - k (T - T_{0})$

$\frac{d T}{(T - T_{0})} = - k d t$

$\Rightarrow k = - \frac{1}{6} l n (\frac{2}{3})$ - (1)

Now $\frac{d T}{d t} = - k (T - T_{0})$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Physics Thermodynamics

A monoatomic gas performs a work of $\frac{Q}{4}$ where Q is the heat supplied to it. The molar heat capacity of the gas will be ________ R during this transformation. Where R is the gas constant.

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Payal Gupta

Contributor-Level 10

$Q = Δ υ + w \Rightarrow n C Δ T = n C_{v} Δ T = \frac{n C Δ T}{4} = \frac{3}{4} n C Δ T = n C_{v} Δ T$

$C = \frac{4}{3} C_{v} = \frac{4}{3} * \frac{3}{2} R = 2 R$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

A circular coil of 1000 turns each with area 1m² is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07 T. The maximum voltage generation will be ________V.

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Payal Gupta

Contributor-Level 10

given N = 1000 turns

A = 1m²

$ω = 1 r e v / s e c$

$= 1 * 2 π r a d / s e c$

$V = \frac{- d ?_{B}}{d t} = \frac{- d}{d t} (N B A c o s ω t)$

= $140 * \frac{22}{7}$

= 20 * 22

= 440 volts

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

A ray of light is incident at an angle of incidence $60 °$ on the glass slab of refractive index $\sqrt[]{3} .$ After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $4 \sqrt[]{3}$ cm. The thickness of the glass slab is _______cm.

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Payal Gupta

Contributor-Level 10

$d = 4 \sqrt[]{3}$ cm (Lateral shift)

By Snell's law

$μ_{a i r} s i n 60 ° = μ_{g} s i n θ$

$\Rightarrow 1 * \frac{\sqrt[]{3}}{2} = \sqrt[]{3} s i n θ$

$s i n θ = \frac{1}{2}$

= 30°

$t a n 30 ° = \frac{d}{t} = \frac{4 \sqrt[]{3}}{t}$

$\Rightarrow \frac{1}{\sqrt[]{3}} = \frac{4 \sqrt[]{3}}{t}$

t = 12 cm

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

A sample contains 10^-2kg each of two substances A and B with half lives 4s and 8s respectively. The ration of their atomic weights is 1 : 2. The ratio of the amounts of A and B after16s is $\frac{x}{100}$ . The value of x is ______.

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Payal Gupta

Contributor-Level 10

$f o r A, t_{\frac{1}{2}} = 4 s e c$

$= m A_{0} e^{-} \frac{0.693}{4} * 16$

$m_{A} = m_{A 0} e^{- 4 * 0.693}$ - (1)

for B,

for B, $t_{\frac{1}{2}} = 8 s e c$

$m_{B} = m_{B_{0}} e^{- 2 * 0.693}$ - (2)

$\frac{m_{A}}{m_{B}} = \frac{m_{A O}}{m_{B O}} \frac{e^{- 4} * 0.693}{e^{- 2} * 0.693}$

$\frac{m_{A}}{m_{B}} = \frac{25}{100} = \frac{x}{100} x = 25$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

In the given circuit, the value of current I_Lwill be _________mA. (When R_L = 1 k $Ω$ )

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Payal Gupta

Contributor-Level 10

$I_{L} = \frac{V}{R_{L}} = \frac{5}{1 k Ω}$

$l_{L} = 5 * 1 0^{- 3} A = 5 m A$

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

Two travelling waves of equal amplitudes and frequencies move in opposite directions along a string. They interfere to produce a stationary wave whose equation is given by y = $(10 c o s π x s i n \frac{2 π t}{T}) c m$

The amplitude of the particle at x = $\frac{4}{3}$ cm will be _______cm.

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Payal Gupta

Contributor-Level 10

$y = (10 c o s π x s i n \frac{2 π t}{T}) c m$

$a t x = \frac{4}{3}$

$y = 10 c o s (\frac{4 π}{3}) s i n (\frac{2 π t}{T}) c m$

$= - 5 s i n (\frac{2 π t}{T}) c m$

Then Amplitude will be 5 cm.

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

A potentiometer wire of length 10m and resistance 20 $Ω$ is connected in series with a 25 V battery and an external resistance $30 Ω$ . A cell of emf E in secondary circuit is balanced by 250cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10} .$ The value of x is ______.

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Payal Gupta

Contributor-Level 10

AB = 10 m

$R_{A B} = 20 Ω$

$L_{A C} = 250 c m$

= 2.5 m

$l = \frac{25}{20 + 30}$

$E = \frac{x}{10} = \frac{25}{10}$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Three Class 12th

An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of ________ GHz.

(Given µ_r = 1 for dielectric medium)

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Payal Gupta

Contributor-Level 10

$v = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8}}{\sqrt[]{6.25 * 1}} = \frac{3 * 1 0^{8}}{2.5} = 1.25 * 1 0^{8} m / s e c$

$A s f λ = f (5 * 1 0^{- 3} * 4) = 1.25 * 1 0^{8} \Rightarrow f = 6.25 G H z$

So, $f \approx 6 .$

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2 months ago

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

A body is projected from the ground at an angle of $45 °$ with horizontal. Its velocity after 2s is 20 ms^-1. The maximum height reached by the body during its motion is ______m. (use g = 10ms^-2)

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Payal Gupta

Contributor-Level 10

after 2 sec, v = 20 m/sec

$\begin{array}{l} \vec{v} = u c o s 45 ° \hat{i} + (u s i n 45 ° - g t) \hat{j} \\ v = 20 = \sqrt[]{{(u c o s 45 °)}^{2} + {(u s i n 45 ° - 20)}^{2}} \end{array}$

$\Rightarrow 400 = u^{2} + 400 - 40 u s i n 45 °$

u = 40 sin 45°

$H_{m a x} = \frac{u^{2} s i n^{2} 45}{2 g} = \frac{4 0^{2} s i n^{2} 45 ° * s i n^{2} 45}{2 g}$

$= \frac{40 * 40}{2 * 10} * \frac{1}{2} * \frac{1}{2} = 20 m$

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