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11 months ago

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

An ice cube of dimensions 60cm * 50cm * 20cm is placed in an insulation box of wall thickness 1cm. The box keeping the ice cube at $0 ° C$ 898

of temperature is brought to a room of temperature $40 ° C .$ The rate of melting of ice is approximately:

(Latent heat of fusion of ice is 3.4 * 105 J kg1 and thermal conducting of insulation wall is 0.05 $W m^{- 1} ° C^{- 1}$ )

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alok kumar singh

Contributor-Level 10

$A r e a = 2 * [l b + b h + h l]$

$\Rightarrow A = 2 * [0.6 * 0.5 + 0.5 * 0.2 + 0.2 * 0.6]$

= 1.04 M2

$R_{t h e r m a l} = \frac{t}{K A} = \frac{1 * 1 0^{- 2}}{0.05 * 1.04}$

$\Rightarrow m = 61 * 1 0^{- 5} k g / s$

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11 months ago

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

As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant 2Nm^‑1. If both are given velocity $υ$ in opposite directions, then maximum elongation of the spring is:

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Vishal Baghel

Contributor-Level 10

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_{r} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{m m}{m + m} = \frac{m}{2}$

By wet

$w_{S P} = Δ K . E .$

$- \frac{1}{2} k x^{2} = 0 - \frac{1}{2} (\frac{m}{2}) {(2 v)}^{2}$

$\frac{2}{2} x^{2} = 0.25 v^{2}$

$x^{2} = 0.25 v^{2}$

$x = \frac{v}{2}$

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11 months ago

Class 12th Thermodynamics

1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8} .$ The value of x is

[R = universal gas constant]

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Payal Gupta

Contributor-Level 10

$Δ Q = Δ U + Δ W$

$\Rightarrow Q = Δ U + \frac{Q}{5} \Rightarrow Δ U = \frac{4 Q}{5} = n C_{v} Δ T \Rightarrow \frac{4 Q}{5} = \frac{5 R}{2} Δ T \Rightarrow Δ T = \frac{8 Q}{25 R}$

$Q = n c Δ T = 1 * C * \frac{80}{25 R} \Rightarrow C = \frac{25 R}{8} \Rightarrow x = 25$

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11 months ago

Class 12th Communication Systems

If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are

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Payal Gupta

Contributor-Level 10

Band Width = 2 * n * the highest modulation frequency

$\Rightarrow n = \frac{90 k H_{z}}{2 * 5 k H z} = 9$

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11 months ago

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

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be:

(Radius of earth will, be:

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Vishal Baghel

Contributor-Level 10

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 * \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} * 100 = 1 %$

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11 months ago

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

A nucleus of mass M at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2 M'}{2} (M' < M),$ The ratio of de Broglie wavelength of two parts will be:

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alok kumar singh

Contributor-Level 10

$\frac{λ_{1}}{λ_{2}} = \frac{h / P_{1}}{h / P_{2}} = \frac{1}{1}$

P1 = P2 as Fnet = 0

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11 months ago

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

A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given π = 3.14)

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Vishal Baghel

Contributor-Level 10

$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³ = 729r³

$R = (729) \frac{1}{3} (r) (\frac{1}{3})$

R = 9r

$Δ u = T (4 π r^{2}) * 729 - T * 4 π R^{2}$

$= T * 4 π * 8 R^{2}$

$\begin{array}{l} = 75.39 * 1 0^{- 5} J \\ Δ u = 7.5 * 1 0^{- 4} J \end{array}$

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11 months ago

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

The maximum and minimum voltage of an amplitude modulated signal are 60V and 20V respectively. The percentage modulation index will be:

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alok kumar singh

Contributor-Level 10

Percentage modulation = $\frac{(V_{m a x} - V_{m i n})}{(V_{m a x} + V_{m i n})} * 100 %$

Percentage modulation = $(\frac{60 - 20}{80 + 20}) * 100$

= 50%

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11 months ago

Class 12th Ray Optics and Optical Instruments

A point source of light S, placed at a distance 60cm infront of the centre of a plane mirror of width 50cm, hangs vertically on a wall. A man walks inform of the mirror along a line parallel to the mirror at a distance 1.2m from it (see in the figure). The distance between the extreme points where he can the image of the light source in the mirror is………….cm

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Payal Gupta

Contributor-Level 10

Maximum Distance = 3 (x + x) = 6x = 3 * 2x = 3 * 50 = 150 cm

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11 months ago

Class 12th Gravitation

A body is projected vertically upward from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:

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alok kumar singh

Contributor-Level 10

$V_{C} = \sqrt[]{\frac{2 G M}{R}}$

Using conservation of Mechanical Energy

$\Rightarrow \frac{- G M m}{R} + \frac{1}{2} * m (\frac{V e^{2}}{9}) = \frac{- G M m}{(R + h)}$

$\Rightarrow \frac{1}{R + h} = \frac{8}{9 R} \Rightarrow h = \frac{R}{8} = \frac{6400}{8} = 800 k m$

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