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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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a year 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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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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a year 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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a year 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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a year 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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a year 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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a year 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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Physics Class 12th

A charge particle is moving in uniform magnetic field $(2 \hat{i} + 3 \hat{j})$ T. If it has an acceleration of $(α \hat{i} - 4 \hat{j}) m / s^{2}$ , then the value of will be

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Payal Gupta

Contributor-Level 10

$\vec{a} ⊥ \vec{B}$

$\Rightarrow \vec{a} . \vec{B} = 0$

$\Rightarrow (2 α - 12) = 0$

= 6

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a year ago

Physics Physics Alternating Current

In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L = X_C is put in series, the power factor becomes P₂. The ratio $\frac{P_{1}}{P_{2}}$ is

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Payal Gupta

Contributor-Level 10

$P_{1} = c o s ? = \frac{R}{z} = \frac{R}{\sqrt[]{X_{L}^{2} + R^{2}}}$

$P_{1} = c o s ? = \frac{R}{R \sqrt[]{2}} = \frac{1}{\sqrt[]{2}}$

So, $\frac{P_{1}}{P_{2}} =$ $1$

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