Physics

Physics Physics Motion in Plane

Contributor-Level 10

For ascent

$t_{a} = \sqrt[]{\frac{2 l}{a_{a s c e n t}}} = \sqrt[]{\frac{2 l}{g (s i n θ + μ c o s θ)}}$

For descent

$t_{d} = \sqrt[]{\frac{2 l}{a_{d e s c e n t}}} = \sqrt[]{\frac{2 l}{g (s i n θ - μ c o s θ)}}$

According question, we can write

$\Rightarrow μ = \frac{3}{5} t a n θ = \frac{3}{5} t a n 30 ° = \frac{\sqrt[]{3}}{5}$

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A butterfly is flying with a velocity $4 \sqrt[]{2} m / s$ in North – East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is:

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Contributor-Level 10

${\vec{v}}_{B W} = 4 \sqrt[]{2} (c o s 45 ° \hat{i} + s i n 45 ° \hat{j}) = 4 \hat{i} + 4 \hat{j}, a n d {\vec{v}}_{W g} = 0 \hat{i} - \hat{j}$

${\vec{v}}_{B g} = {\vec{v}}_{B W} + {\vec{v}}_{W g} = (4 \hat{i} + 4 \hat{j}) + (0 \hat{i} - \hat{j}) = 4 \hat{i} + 3 \hat{j}$

$| {\vec{v}}_{B g} | = 5 m / s \Rightarrow S p e e d o f B u t t e r f l y$

Magnitude of displacement of Butterfly = $| {\vec{v}}_{B g} | * t = 5 * 3 = 15 m$

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In the given figure switches S₁ and S₂ are in open condition. The resistance across ab when the switches S₁ and S₂ are closed is__________ $Ω .$

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Contributor-Level 10

Here

$R_{e q} = (12 / /^{l} 6) + (4 / /^{l} 4) + (6 / /^{l} 12)$

$= (\frac{12 * 6}{18}) + (\frac{4 * 4}{8}) + (\frac{6 * 12}{18})$

= 4 +2 + 4 = 10 $Ω$

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Physics Alternating Current

AC voltage V(t) = 20sin $ω t$ of frequency 50Hz is applied to a parallel plate capacitor. The separation between the plates is 2mm and the area is 1m². The amplitude of the oscillating displacement current for the applied Ac voltage is__________.

[Take $ε_{0} = 8.85 * 1 0^{- 12} F / m]$

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Contributor-Level 10

$C = \frac{ε_{0} A}{d} \Rightarrow X_{C} = \frac{1}{C ω} = \frac{d}{ε_{0} A ω} \Rightarrow l_{0} = \frac{V_{0}}{X_{C}} = \frac{V_{0} ε_{0} A ω}{d} = \frac{2 π f V_{0} ε_{0} A}{d}$

$\Rightarrow l_{0} = \frac{2 * 3.14 * 50 * 20 * 8.85 * 1 0^{- 12} * 1}{2 * 1 0^{- 3}} = 27.79 * 1 0^{- 6} A = 27.79 μ A$

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Physics Thermodynamics

One mole of an ideal gas at 27°C is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be________* 10^-1 J.

[Given : R = 8.3J/mole K, $l n 2$ = 0.6931]

(Round off to the nearest integer)

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Physics Physics Mechanical Properties of Solids

Contributor-Level 10

Since process is isothermal, so

$W = n R T l n (\frac{V_{f}}{V_{i}}) = 1 * 8.3 * 300 * l n (\frac{4}{2}) = 1 * 8.3 * 300 * l n (2)$

$\Rightarrow W = 1 * 8.3 * 300 * 0.6931 \approx 1725.82 J \approx 17258 * 1 0^{- 1} J$

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The value of tension in a long thin metal wire has been changed from T₁ to T₂. The lengths of the metal wire at two different values of tension T₁ and T₂ are $l_{1} a n d l_{2}$ respectively. The actual length of the metal wire is:

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Contributor-Level 10

Assuming Tension in metal wire suspended from roof varies linearly and 0 and T₀ ( developed due its own weight) are tensions are ends of wire of length $l .$ So

$T_{1} α (l_{1} - l), a n d T_{2} α (l_{2} - l)$

$\frac{T_{1}}{T_{2}} = \frac{l_{1} - l}{l_{2} - l} \Rightarrow l = \frac{T_{1} l_{2} - T_{2} l_{1}}{T_{1} - T_{2}}$

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For the forward biased diode characteristics shown in the figure, the dynamic resistance at I_D = 3mA will be_________ $Ω .$

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Contributor-Level 10

$R = \frac{Δ V}{Δ l} = \frac{0.75 - 0.70}{(5 - 3) * 1 0^{- 3}} = \frac{0.05 * 1000}{2} = 25 Ω$

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Physics Class 11th

A body rotating with an angular speed of 600rpm is uniformly accelerated to 1800rpm in 10 sec. The number of rotations made in the process is__________.

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Contributor-Level 10

$?^{2} = ?_{0}^{2} + 2 ? ? ? {(\frac{2 ?}{60} * 1800)}^{2} = {(\frac{2 ?}{60} * 600)}^{2} + 2 (\frac{2 ?}{60} * \frac{1200}{10}) ?$

$? {(60 ?)}^{2} ? {(20 ?)}^{2} = 2 (4 ?) ? ? ? = \frac{80 ? * 40 ?}{2 * 4 ?} = 400 ? ? ? r a d$

Number of rotations = $\frac{?}{2 ?} = \frac{400 ?}{2 ?} = 200$

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A current of 5A is passing through a non-linear magnesium wire of cross – section 0.04m². At every point the direction of current density is at an angle of 60° ;with unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is: (Resistivity of magnesium $ρ = 44 * 1 0^{- 8} Ω m)$

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Contributor-Level 10

$l = J A c o s θ \Rightarrow 5 = J * 0.04 * c o s 60 ° \Rightarrow J = \frac{5}{0.02} = 250 A / m^{2}$

$J = \frac{E}{ρ} \Rightarrow E = ρ J = 44 * 1 0^{- 8} * 250 = 11 * 1 0^{- 5} V / m$

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A series LCR circuit of R = $5 Ω,$ L = 20mH and C = 0.5 $μ F$ is connected across an AC supply of 250 V, having variable frequency. The power dissipated at resonance condition is__________* 10² W.

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