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Physics NCERT Exemplar Solutions Class 12th Chapter Six

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Work, Energy and Power

Water falls from a 40m height dam at the rate of 9 * 10⁴ kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is:

(Take g = 10 ms^-2)

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alok kumar singh

Contributor-Level 10

Given : water flow rate = $(\overset{o}{m})$ = 9 * 10⁴kg/hr

= $\frac{9 * 1 0^{4} k g / s e c}{60 * 60}$

Potential energy available on water per unit time

$\Rightarrow \overset{o}{m} g h = \frac{9 * 1 0^{4}}{60 * 60} * 10 * 40$

$\overset{o}{m} g h = 1 0^{4} w a t t$

50% of it converts into electrical energy = 50% of $\overset{o}{m} g h = \frac{1 0^{4}}{2} w a t t$

Let $η$ be the number of bulbs n = 100 = $\frac{1 0^{4}}{2}$

n = 50

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Work, Energy and Power

A block of mass 2kg moving on a horizontal surface with speed of 4 ms^-¹ enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm^-¹. The speed of the block as it just crosses the rough surface will be:

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alok kumar singh

Contributor-Level 10

by, WET $\int F d x = Δ k . 6$

$\int - k x d x = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$\Rightarrow \frac{- k}{2} {[x_{f}^{2} - x_{i}^{2}]}_{0.5}^{1.5} = \frac{1}{2} * 2 [v_{f}^{2} - 16]$ $[? v_{i} = 4 m / s e c]$

$\frac{- 12}{2} [1 . 5^{2} - 0 . 5^{2}] = [v_{f}^{2} - 16]$

$v_{f}^{2} = 16 - 12 = 4$

vf = 2m/sec

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Dual Nature of Radiation and Matter

An ideal fluid of density 800kgm^-3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x =________. (Given g = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

$\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁-(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Thermodynamics

A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.

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Payal Gupta

Contributor-Level 10

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$= (l_{L} + l_{C})$

Therefore current through R circuit at resonance will be zero

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electromagnetic Induction

As per the given circuit, the value of current through the battery will be__________A.

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Payal Gupta

Contributor-Level 10

$\begin{array}{l} D_{1} \\ D_{3} \end{array}}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I = \frac{v}{R} = \frac{10}{10} = 1 A m p$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Oscillations

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________* 10^-2mm.

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Payal Gupta

Contributor-Level 10

Let 'x' he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

10 y = 9 x

$y = \frac{9 x}{10}$

Least count = $x - \frac{9 x}{10} = \frac{x}{10} = \frac{0.05}{10}$

= 0.005 cm

= 5 * 10^-2 mm

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six 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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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electric Charges and Fields

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Payal Gupta

Contributor-Level 10

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i0 $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$

$= \frac{L i_{0}}{200 * 1 0^{- 6} s e c} = \frac{20 * 1 0^{- 3} * 2}{200 * 1 0^{- 6}} = 400$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Nuclei

The elongation of a wire on the surface of the earth is 10^-⁴m. The same wire of same dimensions is elongated by 6 * 10^-⁵ m on another planet. The acceleration due to gravity on the planet will be__________ms^-². (Take acceleration due to gravity on the surface of earth = 10 ms^-²)

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Payal Gupta

Contributor-Level 10

$Δ L$ (change in length) $= \frac{F l}{A E} = \frac{m g l}{A E}$

$Δ l \propto g$

$\frac{Δ l_{1}}{g_{1}} = \frac{Δ l_{2}}{g_{2}} \Rightarrow \frac{1 0^{- 4}}{10} = \frac{6 * 1 0^{- 5}}{g_{1}}$

$g_{1} = \frac{6 * 1 0^{- 4}}{1 0^{- 4}} =$ 6 m/sec²

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Laws of Motion

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

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Payal Gupta

Contributor-Level 10

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h = u t + \frac{1}{2} a t^{2}$

$= 0 * 1 + \frac{1}{2} * 10 * 1 * 1$

= 5m

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