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11 months ago

Physics Work, Energy and Power

A bag of sand of mass 9.8kg is suspended by a rope. A bullet of 200g travelling with speed 10 ms^-1 gets embedded in it, then loss of kinetic energy will be

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Payal Gupta

Contributor-Level 10

$v = \frac{m v_{0}}{M + m} \Rightarrow K_{f} = \frac{1}{2} (M + m) v^{2} = \frac{m v_{0}^{2}}{2 (M + m)}$

$K_{i} = \frac{m v_{0}^{2}}{2}$

Loss in Kinetic energy = $Δ K = K_{i} - K_{f}$

$Δ K = \frac{1}{2} m v_{0}^{2} (\frac{M}{M + m}) = \frac{1}{2} * 0.2 * 100 = [\frac{9.8}{10}] = 9.8 J$

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11 months ago

Physics Physics Gravitation

An object is taken to a height above the surface of earth at a distance $\frac{5}{4} R$ from the centre of the earth. Where radius of earth, R = 6400km. The percentage decrease in the weight of the object will be

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Payal Gupta

Contributor-Level 10

As we know that $g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{\frac{5}{4} R})}^{2} = \frac{16 g}{25}$

$\Rightarrow Δ g = g - g_{P} = \frac{9 g}{25}$

The percentage decrease in the weight of the object = $\frac{Δ g}{g} * 100 = 36 %$

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11 months ago

Physics Class 11th

Let $η_{1}$ is the efficiency of an engine at T₁ = $447 ° C$ and T₂ = $147 ° C$ while $η_{2}$ is the efficiency at T₁ = $947 ° C$ and T₂ $47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$ will be

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Payal Gupta

Contributor-Level 10

As we know that

$η = 1 - \frac{T_{2}}{T_{1}}, s o$

$\frac{n_{1}}{n_{2}} = \frac{0.42}{0.74} \approx 0.56$

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New answer posted

11 months ago

Physics Class 11th

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is $\sqrt[]{2}$ time the speed of sound, then the value of n will be

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Payal Gupta

Contributor-Level 10

As we know that

$v = \sqrt[]{\frac{γ R T}{M}} \Rightarrow$ Speed of sound in a gas, and

$v_{r m s} = \sqrt[]{\frac{3 R T}{M}} \Rightarrow$ Rms speed of gas molecules, so

$v_{r m s} = \sqrt[]{2} v \Rightarrow \sqrt[]{\frac{3 R T}{M}} = \sqrt[]{2} \sqrt[]{\frac{γ R T}{M}} \Rightarrow γ = \frac{3}{2}$

According formula of specific heat ratio of mixture, we can write

$\Rightarrow$ 5n = 14 n = 2.8

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11 months ago

Physics Physics Oscillations

The length of a seconds pendulum at a height h = 2R from earth surface will be:

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $π^{2} m s^{- 2}$ )

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Payal Gupta

Contributor-Level 10

As we know that

$g_{P} = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}, s o$

$T = 2 π \sqrt[]{\frac{l}{g_{P}}} \Rightarrow l = \frac{g_{p} T^{2}}{4 π^{2}} = \frac{1}{9} m$

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11 months ago

Physics Class 12th

A current of 15 mA flow in the circuit as shown in figure. The value of potential difference between the pointes A and B will be

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Payal Gupta

Contributor-Level 10

$R_{A B} = 5 + (10 / / 5) + 10 = 15 + \frac{10 * 5}{10 + 5} = 15 + \frac{10}{3} = \frac{55}{3} k Ω$

$V_{A B} = R_{A B} l = (\frac{55}{3} * 1 0^{3}) * (15 * 1 0^{- 3}) = 275 V$

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11 months ago

Physics Physics Laws of Motion

A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is:

[Take g = 10m/s^-2]

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alok kumar singh

Contributor-Level 10

$a = μ g = 0.4 * 10 = 4 m / s^{2}$

Slipping stops when V_bag = V_conveyar

0 + at = 2

t = $\frac{2}{a} = \frac{2}{4} = . 5 s e c$ $\frac{}{} \frac{}{}$

So, x = $\frac{1}{2} a t^{2} = \frac{1}{2} * 4 * \frac{1}{4} = 0.5 m$

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11 months ago

Physics Physics Motion in Straight Line

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs^-1 on a conveyor belt moving with a velocity of 5ms^-1. The power needed to keep the belt moving with the same velocity will be:

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alok kumar singh

Contributor-Level 10

$P = \vec{F} . \vec{v} = v \frac{d P}{d t} = v^{2} (\frac{d m}{d t}) = 0.5 * 25 = 12.5 W$

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11 months ago

Physics Units and Measurement

A torque meter is calibrated to reference standards of mass, length and time ach with 5% accuracy. After calibration, the measured torque with this torque meter will have net accuracy of:

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alok kumar singh

Contributor-Level 10

$T o r q u e \vec{τ} = \vec{r} * \vec{F}$

$[τ] = [L] [M L T^{- 2}]$

$[τ] = [M L^{2} T^{- 2}]$

as, $τ = m l^{2} t^{- 2}$

$\frac{d τ}{τ} * 100 = | \frac{d m}{m} * 100 | + 2 | \frac{d l}{l} * 100 | + 2 | \frac{d t}{t} * 100 |$

= 5 + 2 * 5 + 2 * 5

= 25 %

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New answer posted

11 months ago

Physics Physics Moving Charges and Magnetism

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 * 10^-4 Wbm^-2. The frequency of revolution of the electron will be

(Take mass of electron = 9.0 * 10^-31kg)

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Payal Gupta

Contributor-Level 10

f = $\frac{q B}{2 π m} = \frac{1.6 * 1 0^{- 19} * 1.0 * 1 0^{- 4}}{2 * 3.14 * 9 * 1 0^{- 31}} = 0.028 * 1 0^{8} H z = 2.8 * 1 0^{6} H z$

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