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9 months ago

Physics Gravitation

In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA: AB will be x : y. The value of x is

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Payal Gupta

Contributor-Level 10

$\frac{G M}{{(3 R / 2)}^{2}} = \frac{G M}{R^{3}} * r$

$\Rightarrow O A = \frac{4 R}{9} = r$

$A B = R - \frac{4 R}{9} = \frac{5 R}{9} \Rightarrow O A : A B = 4 : 5 = x : y \Rightarrow x = 4$

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9 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Vishal Baghel

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 Ã— 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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9 months ago

Physics Electrostatic Potential and Capacitance

27 similar drops of mercury are maintained at 10V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is……… times that of a smaller drop.

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Payal Gupta

Contributor-Level 10

With the help of conservation of volume, we can write

$27 * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R = 3 r . . . . . . . (1)$

With the help of conservation of charge, we can write

Q = 27 q. (2)

Potential energy of single drop = U1 = $\frac{q^{2}}{8 π ε_{0} r}$

Potential energy of bigger drop = $U_{2} = \frac{Q^{2}}{8 π ε_{0} R} = \frac{27 * 27 * q^{2}}{8 π ε_{0} (3 r)} = 243 (\frac{q^{2}}{8 π ε_{0} r}) = 243 U_{1}$

$\Rightarrow \frac{U_{2}}{U_{1}} = 243$

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9 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Four

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:

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Vishal Baghel

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow \frac{u_{1}}{u_{2}} = \sqrt[]{2}$

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9 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

A transverse wave s represented by y = $2 s i n (ω t - k x) c m .$ The value of wavelength (in cm) for which the wave velocity becomes equal to the maximum particle velocity. Will be:

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alok kumar singh

Contributor-Level 10

Maximum particle velocity = $A ω$

Wave velocity = $\frac{ω}{R}$

$\Rightarrow \frac{ω}{k} = A ω$

$\Rightarrow k = \frac{1}{A} = \frac{1}{2} = c m$

$λ = \frac{2 π}{k} = 4 π c m$

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9 months ago

Physics 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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9 months ago

Physics 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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9 months ago

Physics 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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9 months ago

Physics 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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9 months ago

Physics 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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