Physics

Contributor-Level 10

y₁ = y₂

=> $35 t - \frac{1}{2} g t^{2} = 35 (t - 3) - \frac{1}{2} g {(t - 3)}^{2}$

=> 35 * 3 = $\frac{1}{2} g (t + t - 3) * 3$

105 = $\frac{10}{2} * 3 * (2 t - 3)$

t = 5

$y_{1} = 35 * 5 - \frac{1}{2} * 10 * 5^{2}$

= 50 m

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5 months ago

A parallel-plate capacitor with plate area A has separation d between the plates. Two dielectric slabs of dielectric constant K₁ and K₂ of same area A/2 and thickness d/2 are inserted in the space between the plates. The capacitance of the capacitor will be given by:

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Contributor-Level 10

$C_{e q} = C + \frac{C_{1} . C_{2}}{C_{1} + C_{2}}$

$= \frac{ε_{0} \frac{A}{2}}{d} + \frac{\frac{k_{1} ε_{0} \frac{A}{2}}{\frac{d}{2}} . \frac{k_{2} ε_{0} \frac{A}{2}}{\frac{d}{2}}}{\frac{k_{1} ε_{0} \frac{A}{2}}{\frac{d}{2}} + \frac{k_{2} ε_{0} \frac{A}{2}}{\frac{d}{2}}}$

= $\frac{ε_{0} A}{d} (\frac{1}{2} + \frac{k_{1} k_{2}}{k_{1} + k_{2}})$

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5 months ago

What equal length of an iron wire and a copper-nickel alloy wire, each of 2mm diameter connected parallel to give an equivalent resistance of 3 $Ω$ ?

(Given resistivities of iron and copper-nickel alloy wire are $12 μ Ω$ cm and 51 $μ Ω$ cm respectively)

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Contributor-Level 10

$\frac{R_{1} R_{2}}{R_{1} + R_{2}} = 3$

$\Rightarrow \frac{(ρ_{1} \frac{l}{A}) (ρ_{2} \frac{l}{A})}{(\frac{ρ_{1} l}{A}) + (\frac{ρ_{2} l}{A})} = 3$

$= \frac{3 * [\frac{π}{4} * {(2 * 1 0^{- 3})}^{2}] [12 + 51] * 1 0^{- 6} * 1 0^{- 2}}{12 * 1 0^{- 6} * 1 0^{- 2} * 51 * 1 0^{- 6} * 1 0^{- 2}}$

$= 97$

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5 months ago

A transmitting antenna at top of a tower has a height of 50m and the height of receiving antenna is 80m. What is the range of communication for Line of Sight (LoS) mode? [use radius of earth = 6400km]

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Contributor-Level 10

Range of communication

= d₁ + d₂

= $\sqrt[]{2 R h_{T}} + \sqrt[]{2 R h_{R}}$

= $\sqrt[]{2 * 6400} (\sqrt[]{0.05} + \sqrt[]{0.08})$

= 57.28 km

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5 months ago

Statement-I: By doping silicon semiconductor with pentavalent material, the electrons density increases.

Statement-II: The n-type semiconductor has net negative charge.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Physics Physics Mechanical Properties of Solids

Contributor-Level 10

Pentavalent materials have more electrons and so electron density increase. But overall semiconductor is neutral

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5 months ago

Two blocks of masses 3kg and 5kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{π} * 1 0^{2} N m^{- 2} .$ What is the minimum radius of the wire? (take g = 10ms^-2)

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Contributor-Level 10

Tension in wire, $T = \frac{2 * 3 * 5 * 10}{3 + 5} = \frac{75}{2} N$

$\frac{T}{π r_{m i n}^{2}} = \frac{24}{π} * 1 0^{2}$

$\Rightarrow r_{m i n}^{2} = \frac{75}{2 * 24} * 1 0^{- 2} = \frac{25}{16} * 1 0^{- 2}$

$\Rightarrow r_{m i n} = \frac{5}{4} * 1 0^{- 1} m = 12.5 c m$

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Physics Class 11th

The initial mass of a rocket is 1000kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20ms^-². The gases come out at a relative speed of 500ms^-¹ with respect to the rocket: [Use g = 10m/s²]

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Contributor-Level 10

$V_{r e l} \frac{d m}{d t} - m g = F_{e x t}$

=>500 $\frac{d m}{d t} - 1000 * 10 = 1000 * 20$

$\frac{d m}{d t} = 60 k g / s$

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5 months ago

Identify the logic operation carried out by the given circuit:-

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Physics Mechanical Properties of Fluids

Contributor-Level 10

A B X Y Z

1 0 0

1 0 1

...more

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5 months ago

Two narrow bores of diameter 5.0mm and 8.0 mm are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube.

[Take surface tension of water T = 7.3 * 10^-² Nm^-¹, angle of contact = 0, g = 10 ms^-² and density of water = 1.0 * 10³ kg m^-³]

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Physics Physics Oscillations

Contributor-Level 10

P_B – P_A = $ρ g h$

$\Rightarrow (P_{0} - \frac{2 T}{r^{2}}) - (P_{0} - \frac{2 T}{r_{1}}) = ρ g h$

$h = \frac{2 T (\frac{1}{r_{1}} - \frac{1}{r_{2}})}{ρ g}$

$= \frac{2 * 7.3 * 1 0^{- 2} * (\frac{1}{2.5 * 1 0^{- 3}} - \frac{1}{4 * 1 0^{- 3}})}{1000 * 10}$

$\begin{array}{l} = 2.19 * 1 0^{- 3} m \\ = 2.19 m m \end{array}$

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5 months ago

If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is

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