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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 Class 12th

For the given circuit the current through battery of 6 V just after closing the switch 'S' will be_______A.

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Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

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

For the given circuit the current through battery of 6 V just after closing the switch ‘S’ will be_______A.

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

An electrical bulb rated 220 V, 100 W, is connected in series with another bulb rated 220 V, 60 W. If the voltage across combination is 220 V, the power consumed by the 100 W bulb will be about ______W.

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Vishal Baghel

Contributor-Level 10

$\frac{v_{1}}{v_{2}} = \frac{R_{1}}{R_{2}} = \frac{P_{2}}{P_{1}} = \frac{60}{100} = \frac{3}{5}$

$v_{1} = \frac{3}{8} * 220 = 3 * 27.5 = 82.5 v$

$∴ P_{1} = {(\frac{v_{1}}{v})}^{2} P$

= 14 w

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

A battery of $3.0 V$ is connected to a resistor dissipating $0.5 W$ of power. If the terminal voltage of the battery is $2.5 V$ , the power dissipated within the internal resistance is:

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alok kumar singh

Contributor-Level 10

$\begin{array}{r} ↑_{1}^{?} (1) ε = 3 \\ (2) ε - I r = 2.5 V \\ \Rightarrow I r = 0.5 \\ Now, IR = 2.5 \\ \Rightarrow \frac{R}{r} = 5 . \\ \Rightarrow \frac{P_{R}}{R_{r}} = \frac{I^{2} R}{I^{2} r} = \frac{R}{r} = 5 \\ \Rightarrow P_{r} = \frac{0.5}{5} = 0.1 \end{array}$

$266.67$

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

Given below are two statements:

Statement I : A uniform wire of resistance 80 $Ω$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5 $Ω .$

Statement II : Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3 : 2.

In the light of the above statements, choose the most appropriate answer from the option given below

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Vishal Baghel

Contributor-Level 10

Statement – I

$80 = \frac{ρ l}{A}$

$S o, R_{e q} = \frac{R'}{4} = \frac{20}{4} = 5 Ω$

Statement – II

$\frac{P_{1}}{P_{2}} = \frac{V^{2}}{2 R} * \frac{3 R}{V^{2}} = \frac{3}{2}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

Dimensional formula for thermal conductivity is (here $K$ denotes the temperature):

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alok kumar singh

Contributor-Level 10

$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Wheatstone Bridge

Four resistance $40 Ω, 60 Ω, 90 Ω$ , and $110 Ω$ make the arms of a quadrilateral ABCD. Across $A C$ is a battery of emf $40 V$ and internal resistance negligible. The potential difference across $B D$ in $V$ is

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alok kumar singh

Contributor-Level 10

10553.33

Sol. $\frac{1}{λ_{min}} = R [\frac{1}{1} - \frac{1}{\infty}] = R [n = \infty \to n = 1]$

$\frac{1}{λ_{m a x}} = R [\frac{1}{1} - \frac{1}{4}] = \frac{3 R}{4} [n = 2 \to n = 1]$

$\Rightarrow Δ λ \Rightarrow \frac{4}{3 R} - \frac{1}{R} \Rightarrow \frac{1}{3 R} = 340$

For Paschan $\Rightarrow \frac{1}{λ_{min}} = R [\frac{1}{9}] [n = \infty \to h = 3]$

$\Rightarrow \frac{1}{λ_{m a x}} = R [\frac{1}{9} - \frac{1}{16}] = \frac{7 R}{144}$

$\Rightarrow Δ λ = \frac{81}{7 R}$

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

Two metallic wires of identical dimensions are connected in series. If $σ_{1} a n d σ_{2}$ are the conductivities of these wires respectively, the effective conductivity of the combination is:

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Vishal Baghel

Contributor-Level 10

R = R1 + R2

$\Rightarrow \frac{2 l}{σ A} = \frac{l}{σ_{1} A} + \frac{l}{σ_{2} A} \Rightarrow \frac{2}{σ} = \frac{1}{σ_{1}} + \frac{1}{σ_{2}} \Rightarrow σ = \frac{2 σ_{1} σ_{2}}{σ_{1} + σ_{2}}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

The speed verses time graph for a particle is shown in the figure. The distance travelled (in $m$ ) by the particle during the time interval $t = 0$ to $t = 5 s$ will be $\dots .$ .

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alok kumar singh

Contributor-Level 10

Sol.

$\Rightarrow n_{1} T_{1} + n_{2} T_{2} = n T$

$\Rightarrow (0.1) (200) + (0.05) (400) = (0.15) T$

$\Rightarrow T = 266.67$

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Physics NCERT Exemplar Solutions Class 12th Chapter Three Motion in a Straight Line

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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