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Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Nine Ray Optics and Optical Instruments

A light ray is incident, at an incident angle q₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ___________ degree.

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Payal Gupta

Contributor-Level 10

$δ_{t o t a l} = 360 ° - 2 θ$

= 360 – 2 * 75° = 210°

Total deviation = $δ = 150 °$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Mechanical Properties of Solids

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed $υ$ and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by:

(R = universal gas constant)

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Payal Gupta

Contributor-Level 10

$K E = Δ U$

$Δ U = \frac{1}{2} m v^{2}$

$n C_{v} Δ T = \frac{m v^{2}}{2}$

$\Rightarrow \frac{m}{M} \frac{R}{y - 1} Δ T = \frac{m v^{2}}{2}$

$Δ T = \frac{0.4}{2} \frac{M v^{2}}{R}$

$Δ T = \frac{M v^{2}}{5 R}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Ray Optics and Optical Instruments

The efficiency of a Carnot's engine, working between stem point and ice point, will be

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Payal Gupta

Contributor-Level 10

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$η = (1 - \frac{T_{2}}{T_{1}}) * 100 = (1 - \frac{273}{373}) * 100 = 26.81 %$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Ray Optics and Optical Instruments

A parallel beam of light is allowed to afall on a transparent spherical globe of diameter 30cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is_________ mm.

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alok kumar singh

Contributor-Level 10

When object is at infinity

u = $\infty, μ_{1} = 1, R = 15 c m$

v = ? $μ_{2} = 1.5$

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1}{\infty} + \frac{1.5}{v} =$ $\frac{1.5 - 1}{15}$

$\frac{1.5}{v} =$ $\frac{0.5}{15}$

v =

45cm

Now for Refraction through C2

u = + 15cm

v = ?

$μ_{1} = 1.5$

$μ_{2} = 1$

R = 15

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1.5}{15} + \frac{1}{v} = \frac{1 - 1.5}{- 15}$

$\frac{- 1}{10} + \frac{1}{v} = \frac{1}{30}$

$\frac{1}{v} = \frac{1}{30} + \frac{1}{10} = \frac{4}{10}$

$v = \frac{30}{4} = + 7.5 c m$

from centre = 15 + 7.5 = 22.5 cm = 225mm

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Mechanical Properties of Solids

A wire of length L is hanging from a fixed support. The length changes to L₁ andL₂when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

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alok kumar singh

Contributor-Level 10

Spring constant of wire (k) = $\frac{A E}{L}$

Case 1

$k [Δ] = 1 g$

$k [L_{1} - L] = 1 g$ - (1)

Case 2 $k [L_{2} - L] = 2 g$ - (2)

$(1) \div (2)$

$\frac{L_{1} - L}{L_{2} - L} = \frac{1}{2}$

L = 2L₁ – L₂

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Mechanical Properties of Solids

A uniformly heavy rod of mass 20kg, cross sectional area 0.4 m² an length 20m is hanging from a fixe support. Neglecting the lateral contraction. The elongation in the rod due to its own weight is x * 10^-9m. The value of x i________.

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alok kumar singh

Contributor-Level 10

$\int_{0}^{Δ l} d y = \int_{0}^{l} (\frac{m g}{l}) * \frac{d x}{A Y}$

$\Rightarrow Δ l = \frac{m g l}{2 A Y} = 25 * 1 0^{- 9} m$

$\Rightarrow x = 25.00$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Ray Optics and Optical Instruments

The graph between $\frac{1}{u} a n d \frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is 1.5 and its both the surfaces have same radius of curvature R. The value of R will be _________cm. (where u = object distance, v = image distance)

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Vishal Baghel

Contributor-Level 10

For point B, $\frac{1}{u} = - 0.10 c m^{- 1}, \frac{1}{v} = 0$

Thus u = 10 cm, v = $\infty$

i.e., f = 10 cm

$\frac{1}{10} = (1.5 - 1) (\frac{2}{R}) \Rightarrow \frac{1}{R} = \frac{1}{10}$

R = 10 cm

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Ray Optics and Optical Instruments

In the given figure, the face AC of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle $60 °$ at the side AC, the refracteel light beam just grazes along face AC. The refractive index of the liquid $n = \frac{\sqrt[]{x}}{4} .$

The value of x is_______.

(Given refractive index of glass = 1.5)

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alok kumar singh

Contributor-Level 10

1.5 * sin 60 = R * sin 90

$\Rightarrow 1.5 * \frac{\sqrt[]{3}}{2} = n = \frac{\sqrt[]{x}}{4}$

x = 27

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Physics Mechanical Properties of Solids

The area of cross section of the rope used to lift a load by a crane is 2.5 * 10^-4m². The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be:

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alok kumar singh

Contributor-Level 10

Breading stress = $\frac{W e i g h t}{A r e a}$

$\Rightarrow \frac{10}{2.5 * 10.4} = \frac{25}{A}$

$\Rightarrow A = 6.25 * 1 0^{- 4} m^{2}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Nine Mechanical Properties of Solids

In an experiment to determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young's modulus $(Δ Y)$ is found to be x * 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

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Vishal Baghel

Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 10^-3 m

$\frac{F}{A} = y \frac{Δ L}{L}$

$y = \frac{F L}{A Δ L} = \frac{(m g) * 1}{(\frac{π d^{2}}{4}) * 0.4 * 1 0^{- 3}}$

$Δ y = 0.1 * 0.199 * 1 0^{12} = 1.99 * 1 0^{10}$

= 1.99

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