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6 months ago

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

$A B C$ is a plane lamina of the shape of an equilateral triangle. $D, E$ are mid points of $A B, A C$ and $G$ is the centroid of the lamina. Moment of inertia of the lamina about an axis passing through $G$ and (Rotational)

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alok kumar singh

Contributor-Level 10

$\begin{array}{l} A H \to \frac{H}{2} & H = \frac{\sqrt[]{3} a}{2} \\ A G \to \frac{H}{\sqrt[]{3}} & A l = \frac{H}{2 \sqrt[]{3}} \\ M_{A B C} = M & M_{A D E} = \frac{M}{4} \end{array}$

$\Rightarrow I_{G} = \frac{M a^{2}}{12} - [\frac{M}{4} \frac{{[\frac{a}{2}]}^{2}}{12} + \frac{M}{4} {[\frac{a}{2 \sqrt[]{3}}]}^{2}] = (\frac{11}{16}) \frac{{M a}^{2}}{12}$

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6 months ago

Class 12th Application of Integrals

The area of the region S = ${(x, y) : y^{2} \leq 8 x, y \geq \sqrt[]{2} x, x \geq 1}$ is

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Payal Gupta

Contributor-Level 10

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6 months ago

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

Two streams of photons, possessing energies equal to five and ten times the work function of metal are incident on the metal surface successively. The ratio of maximum velocities of the photoelectron emitted, in the two cases respectively, will be

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Vishal Baghel

Contributor-Level 10

$E_{1} = 5 ? - ? = 4 ? = \frac{1}{2} m v_{1}^{2}$

$E_{2} = 10 ? - ? = \frac{1}{2} m V_{2}^{2}$

$\Rightarrow \frac{V_{1}}{V_{2}} = \frac{\sqrt[]{\frac{8 ?}{m}}}{\sqrt[]{\frac{18 ?}{m}}} = \sqrt[]{\frac{8}{18}} = \frac{\sqrt[]{4}}{9} = \frac{2}{3}$

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6 months ago

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

The power of lens (biconvex) is 1.25 m^-¹ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium:

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Vishal Baghel

Contributor-Level 10

Power of a lens, P = 1.25 m^-1

$μ_{l e n s} = 1.5, r_{1} = 20 c m, r_{2} = 40 c m$

As, P = $\frac{1}{f} = 1.25$

$\Rightarrow \frac{1.5}{μ_{1}} = \frac{1}{6} + 1 = \frac{7}{6} \Rightarrow μ_{1} = \frac{6}{7} * 1.5 = \frac{9}{7}$

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6 months ago

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

In a compound microscope, the magnified virtual image is formed at a distance of $25 c m$ from the eye-piece. The focal length of its objective lens is $1 c m$ . If the magnification is 100 and the tube length of the microscope is $20 c m$ , then the focal length of the eyepiece lens (in $c m$ ) is

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alok kumar singh

Contributor-Level 10

$M = \frac{1}{f_{o}} [1 + \frac{D}{F e}]$

$\Rightarrow 100 = \frac{20}{1} [1 + \frac{25}{F e}]$

$\Rightarrow 1 + \frac{25}{F e} = 5$

$\Rightarrow F e = \frac{25}{4} = 6.25 c m$

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6 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter One

Let f : R → R be defined as

where a, b, c $\in$ R and [t] denotes greatest integer less than or equal to t. Then, which of the following statements is true?

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Payal Gupta

Contributor-Level 10

For x < 0 0 < e^x < 1 [e^x] = 0

$0 \leq x < 1 a e^{x} + [x - 1]$

$= a e^{x} + [x] - 1$

= a e^x – 1 b + [sin px]

$∴ f (x) = [\begin{array}{l} 0 x < 0 \\ a e^{x} - 1 0 \leq x < 1 \\ b - 1 1 \leq x < 2 \\ - c x \geq 2 \end{array}]$

For f to be continuous at x = 0

a – 1 = 0 ⇒ a = 1

$\begin{array}{l} ∴ a + b + c = 1 + e + 1 - e = 2 \\ a + b + c \neq 1 \end{array}$

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6 months ago

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

Sun light falls normally on a surface of area 36cm² and exerts an average force of 7.2 * 10^-⁹N within a time period of 20 minutes. Considering a case of complete adsorption, the energy flux of incident light is

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Vishal Baghel

Contributor-Level 10

A = 36 cm²= 36 * 104 m²

$? F ? = 7.2 * 1 0^{- 9} N, t = 20 m i n .$

$? F ? = \frac{Δ P}{Δ t} = \frac{E}{C Δ t} \Rightarrow \frac{E}{Δ t} = ? F ? c \Rightarrow \frac{E}{A Δ t} = \frac{? F ? c}{A}$

= 0.06 W/cm²

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6 months ago

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

A closed vessel contains 0.1 mole of a monatomic ideal gas at $200 K$ . If 0.05 mole of the same gas at $400 K$ is added to it, the final equilibrium temperature (in $K$ ) of the gas in the vessel will be close to

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alok kumar singh

Contributor-Level 10

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

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

$\Rightarrow T = 266.67$

$6.25$

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6 months ago

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

A transformer operating at primary voltage 8 kV and secondary voltage 160V serves a load of 80 k.W. Assuming the transformer to be ideal with purly resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

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Vishal Baghel

Contributor-Level 10

$V_{P} = 8 k V, V_{S} = 160 v P = 80 k W$

(Purely resistive load)

$P = V I = \frac{V_{P}^{2}}{R_{P}} = \frac{V_{S}^{2}}{R_{S}}$

$R_{S} = \frac{V_{S}^{2}}{P} = \frac{160 * 160}{80 * 1 0^{3}} = 320 * 1 0^{- 3} = 0.32 Ω$

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6 months ago

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

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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