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

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

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Vishal Baghel

Contributor-Level 10

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

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Class 12th Oscillations

A particle executes S.H.M. with amplitude â€˜aâ€™ and time period â€˜Tâ€™. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt[]{x} a}{2} .$ The value of x is

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Payal Gupta

Contributor-Level 10

$v = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow \frac{A ω}{2} = ω \sqrt[]{A^{2} - s^{2}} \Rightarrow s = \frac{\sqrt[]{3} A}{2} \Rightarrow x = 3$

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

A gas has n degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be:

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alok kumar singh

Contributor-Level 10

$C v = \frac{n * R}{2}$

$C p = C v + R = \frac{(n + 2) R}{2}$

$\Rightarrow \frac{C v}{C P} = \frac{n}{(n + 2)}$

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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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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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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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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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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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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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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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