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11 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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11 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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11 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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11 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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11 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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11 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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11 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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11 months ago

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

Let [t] denote the greatest integer less than or equal to t. Then, the value of the integral $\int_{0}^{1} [- 8 x^{2} + 6 x - 1] d x$ is equal to

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Payal Gupta

Contributor-Level 10

$\int_{0}^{1} [- 8 x^{2} + 6 x - 1] d x$

Let f (x) = -8x² + 6x – 1

f (0) = -1, f (1) = -3

max f (x) = $- \frac{D}{4 a} = - \frac{36 - 32}{- 32} = \frac{1}{8}$

= $- \frac{1}{4} - 1 (\frac{3}{4} - \frac{λ}{2}) - 2 (\frac{\sqrt[]{17} - 3}{8} - \frac{3}{4}) - 3 (1 - \frac{\sqrt[]{17} - 3}{8})$

= $\frac{\sqrt[]{17} - 13}{8}$

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11 months ago

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

The magnetic field at the centre of current carrying circular loop is B₁. The magnetic field at a distance of $\sqrt[]{3}$ times radius of the given circular loop from the center on its axis is B₂. The value of B₁/B₂ will be

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Vishal Baghel

Contributor-Level 10

$B_{1} = \frac{μ_{0} i}{2 R}$

$B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + {(\sqrt[]{3} R)}^{2})}^{3 / 2}} = \frac{μ_{0} i R^{2}}{2 * 8 R^{3}} = \frac{μ_{0} i}{8 * 2 R} = \frac{B_{1}}{8}$

$\Rightarrow \frac{B_{1}}{B_{2}} = 8 : 1$

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11 months ago

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

Blocks of masses $m, 2 m$ and $8 m$ are arranged in a line on a frictionless floor. Another block of mass $m$ , moving with speed $v$ along the same line (see figure) collides with mass $m$ in perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass $8 m$ starts moving the total energy loss is $p %$ of the original energy. Value of ' $p$ ' is close to:

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alok kumar singh

Contributor-Level 10

$? v$

$\Rightarrow m v = 16 {m v}_{1}$

$\Rightarrow V_{1} = \frac{V}{16}$

$\Rightarrow Δ k$ loss $= \frac{1}{2} {m v}^{2} - \frac{1}{2} (16 m) {(\frac{V}{16})}^{2}$

$\Rightarrow \frac{1}{2} {m v}^{2} (\frac{15}{16})$

$%$ loss $= \frac{15}{16} * 100 = 93.75 %$

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