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11 months ago

Physics Class 11th

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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Payal Gupta

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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11 months ago

Physics Work, Energy and Power

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 in opposite directions, then maximum elongation of the spring is

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Payal Gupta

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

Physics Physics Gravitation

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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Payal Gupta

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

Physics Class 11th

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 p = 3.14)

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Payal Gupta

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

Physics Physics Thermodynamics

A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be

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Payal Gupta

Contributor-Level 10

Constant Entropy means Adiabatic process

$P v^{y} = C$

$P_{1} {v_{1}}^{y} = P_{2} {v_{2}}^{y}$

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P2 = 32 P

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11 months ago

Physics Physics Thermodynamics

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

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Payal Gupta

Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 * \frac{3}{2} * 8.3 * 40$

$= 3486 J$

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11 months ago

Physics Physics Oscillations

When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be

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Payal Gupta

Contributor-Level 10

x = A sin $ω t$

$\Rightarrow v = A ω c o s ω t$

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

$\frac{v^{2}}{ω^{2}} + x^{2} = A^{2}$

Elliptical

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11 months ago

Physics Class 12th

The total charge on the system of capacitors $C_{1} = 1 μ F, C_{2} = 2 μ F, C_{3} = 4 μ F a n d C_{4} = 3 μ F$ connected in parallel is:

(Assume a battery of 20V is connected to the combination)

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Payal Gupta

Contributor-Level 10

$C_{e q} = C_{1} + C_{2} + C_{3} + C_{4}$

= 1 + 2 + 4 + 3

$= 10 μ F$

Q = C_eq v

= (10 * 20) $μ C$

Q = 200 $μ C$

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11 months ago

Physics Class 12th

The current I in the above given circuit will be

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Payal Gupta

Contributor-Level 10

$R = \frac{8}{2} Ω = 4 Ω$ (wheat stone bridge)

$I = \frac{V}{R} = \frac{40}{4} = 10 A$

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11 months ago

Physics Class 12th

B_X and B_Y are the magnetic fields at the centre of two coils X and Y respectively each carrying equal current. If coil X has 200 turns and 20 cm radius and coil Y has 400 turns and 20 cm radius, the ratio of B_X and B_Y is

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Payal Gupta

Contributor-Level 10

$B = N \frac{μ_{0} l}{2 μ}$

$B α \frac{N}{μ}$

$\frac{β_{x}}{β_{y}} = \frac{N_{x}}{r_{x}} * \frac{r_{y}}{N_{y}}$

$\frac{B_{x}}{B_{y}} = \frac{200}{20} * \frac{20}{400} = \frac{1}{2}$

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