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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

A person pushes a box on a rough horizontal platform surface. He applies a force of 200 $N$ over a distance of $15 m$ . Thereafter, he gets progressively tired and his applied force reduces linearly with distance to $100 N$ . The total distance through which the box has been moved is $30 m$ . What is the work done by the person during the total movement of the box?

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alok kumar singh

Contributor-Level 10

Sol. (n) $λ = 5$

$(n, m)$ : Integers

$\begin{array}{r} (\frac{2 m + 1}{2}) λ = \frac{3}{2} \\ \Rightarrow \frac{3 / 2}{5} = \frac{2 m + 1}{2 n} \\ \Rightarrow 3 n = 10 m + 5 \end{array}$

$N, m$ are integers.

So, $m = 1, n = 5, λ = 1$

$\begin{array}{l} m = 4 & n = 15 & λ = \frac{1}{3} \\ m = 7 & n = 25 & λ = \frac{1}{5} \end{array}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eight

The electric field of a plane electromagnetic wave is given by $\overset{?}{E} = E_{0} (\hat{x} + \hat{y}) s i n ? (k z - ω t)$ Its magnetic field will be given by:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

If t = $\sqrt[]{x} + 4, t h e n {(\frac{d x}{d t})}_{t = 4} i s :$

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Vishal Baghel

Contributor-Level 10

$t = \sqrt[]{x} + 4$

then $\frac{d t}{d x} = \frac{1}{2 \sqrt[]{x}} ∴ {(\frac{d x}{d t})}_{t = 4} = {(2 \sqrt[]{x})}_{t = 4} = {[2 (t - 4)]}_{a t t = 4}$

= 0

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Seven

A series $L - R$ circuit is connected to a battery of emf $V$ . If the circuit is switched on at $t =$ 0 , then the time at which the energy stored in the inductor reaches $(1 / n)$ times of its maximum value, is :

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alok kumar singh

Contributor-Level 10

Sol. Mono atomic $? C_{v} = \frac{3 R}{2} C_{p} = \frac{5 R}{2}$

Di- atomic $? C_{V} = \frac{5 R}{2} C_{p} = \frac{7 R}{2}$

(Rigid)

Di-atomic $? C_{v} = \frac{7 R}{2} C_{p} = \frac{9 R}{2}$

(Non-Rigid)

Tri-atomic $? C_{V} = 3 R C_{P} = 4 R$

(Rigid)

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height $\frac{h}{3}$ while going up and coming down respectively.

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Vishal Baghel

Contributor-Level 10

$h = \frac{u^{2}}{2 g} \Rightarrow u = \sqrt[]{2 g h}$

$\frac{h}{3} = \sqrt[]{2 g h} t - 5 t^{2}$

$5 t^{2} - \sqrt[]{20 h} t + \frac{h}{3} = 0$

$t = \frac{\sqrt[]{20 h} \pm \sqrt[]{20 h - 4 * 5 * \frac{h}{3}}}{10}$

$∴ \frac{t_{1}}{t_{2}} = \frac{\sqrt[]{20 h} - \sqrt[]{\frac{40}{3} h}}{\sqrt[]{20 h} + \sqrt[]{\frac{40}{3} h}} = \frac{1 - \sqrt[]{\frac{2}{3}}}{1 + \sqrt[]{\frac{2}{3}}} = \frac{\sqrt[]{3} - \sqrt[]{2}}{\sqrt[]{3} + \sqrt[]{2}}$

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6 months ago

Physics physics ncert exemplar solutions class 12th chapter two

Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is and radius of the drop is r, then

$T = K \sqrt[]{\frac{ρ r^{3}}{S^{\frac{3}{2}}}}$ is dimensionally correct, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

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Vishal Baghel

Contributor-Level 10

$T = k \sqrt[]{\frac{ρ r^{3}}{S^{3 / 2}}}$

$[T]$ ${[M L^{- 3}]}^{1 / 2} {[L^{3}]}^{1 / 2} {[M L T^{- 2} L^{- 1}]}^{\frac{- 3}{4}}$

$[T]$ $= [M^{1 / 2} L^{- \frac{3}{2}} L^{\frac{3}{2}} M^{- \frac{3}{4}} T^{\frac{3}{2}}]$

$= [M^{\frac{1}{4}} T^{3 / 2}]$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$ . The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x 1$ and in the other, $x 2$ . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

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alok kumar singh

Contributor-Level 10

Sol. $U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

The value of current $i_{1}$ flowing from $A$ to $C$ in the circuit diagram is:

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alok kumar singh

Contributor-Level 10

Sol. $\overset{?}{τ} = \overset{?}{μ} * \overset{?}{B}$

$\begin{array}{l} \Rightarrow & 0.018 = μ (0.06) (s i n ? 30^{?}) \\ \Rightarrow & μ = 0.6 \\ \Rightarrow & Work = U_{f} - U_{i} \\ \Rightarrow & 2 μ B \\ \Rightarrow & 7.2 * 10^{- 2} J . \end{array}$

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6 months ago

Physics physics ncert exemplar solution class 12th chapter one

A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E_{0} (1 - a x^{2})$ in the $x$ -direction, where $a$ and $E_{0}$ are constants. Initially the particle was at rest at $x = 0$ . Other then the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is:

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alok kumar singh

Contributor-Level 10

Sol. $E = (I) (t) (A) ?{c o s}^{2} ? θ?$
$\Rightarrow (3.3) [\frac{2 π}{31.4}] [3 * 10^{- 4}] * \frac{1}{2}$

$\Rightarrow 0.99 * 10^{- 4}$

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6 months ago

Physics Semiconductor Electronics

The cut-off voltage of the diodes (shown in figure) in forward bias is 0.6 V. The current through the resister of 40 $Ω$ is………….mA

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Payal Gupta

Contributor-Level 10

D₁ → forward biased

D₂ → Reversed biased

So, current will flow through only D₁

$l = \frac{(1 - 0.6) v}{(60 + 40) Ω} = \frac{0.4}{100}$

= 4 * 10^-³ A

= 4mA

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