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Physics Class 11th

A block of mass 40kg slides over a surface, when a mass of 4kg is suspended through an inextensible massless string passing over frictionless pulley as shown below.

The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (given g = 10 ms^-².)

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Payal Gupta

Contributor-Level 10

According to Newton's laws of motion, we can write

4a = 4g – T . (1), and

40a = T - fk = T $μ$ (40g) ……………… (2)

Adding equations (1) and (2), we can write

44a = 40 – 0.02 * (400) = 32

$\Rightarrow a = \frac{32}{44} = \frac{8}{11} m / s^{2}$

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New answer posted

a year ago

Physics Class 12th

Given below are two statements:

Statement I : The electric force changes the speed of the charged particle and hence charges its kinetic energy; whereas the magnetic force does not change like kinetic energy of the charged particle.

Statement II : The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerated the moving charged particle along the direction of magnetic field.

In the light of the above statements, chose the most appropriate answer from the option given below

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Payal Gupta

Contributor-Level 10

According to Lorentz's Force, we can write

$\vec{F} = {\vec{F}}_{E l e c t r i c} + {\vec{F}}_{M a g n e t i c} = q \vec{E} + q (\vec{v} * \vec{B}), s o$

Statement I is correct but Statement II is incorrect

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New answer posted

a year ago

Physics Class 11th

A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is

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Payal Gupta

Contributor-Level 10

According to question, we can write

Total moles of gas = n = n_Oxygen+ n_Oxygen = $\frac{16}{2} + \frac{128}{32} = 12 m o l e s$

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵cm3 $\approx 27 * 1 0^{4} c m^{3}$

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New answer posted

a year ago

Physics Class 11th

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y = A s i n (π t + ?),$ where time is measured in second. The length of pendulum is

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Payal Gupta

Contributor-Level 10

According to question, we can write

$ω = π = \sqrt[]{\frac{g}{l}} \Rightarrow l = \frac{g}{π^{2}} = \frac{9.8}{{(3.14)}^{2}} = 0.99395 = 99.4 c m$

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New answer posted

a year ago

Physics Physics Gravitation

The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by

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Payal Gupta

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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New answer posted

a year ago

Physics Class 11th

The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three its previous value, then approximate new time period of the satellite will be

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Payal Gupta

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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New answer posted

a year ago

Physics Physics Moving Charges and Magnetism

Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 $μ T$ . the equal current flowing in the two conductors is

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Payal Gupta

Contributor-Level 10

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_{P} = \frac{μ_{0} I}{2 π r} * 2$

$\Rightarrow 300 * 1 0^{- 6} = \frac{4 π * 1 0^{- 7}}{2 π * 4 * 1 0^{- 2}} * 2$

$\Rightarrow$ I = 30 A

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New answer posted

a year ago

Physics Class 11th

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Payal Gupta

Contributor-Level 10

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

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New answer posted

a year ago

Physics Class 12th

A capacitor is discharging through a resistor R. Consider in time t₁ the energy stored in the capacitor reduces to half of its initial value and in time t₂, the charge stored in the capacitor reduces to half of its initial value and in time t₂ the charge stored reduces to one eight of its initial value. The ratio t₁/t₂ will be

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Payal Gupta

Contributor-Level 10

According to Law of discharging of capacitor, we can write

$Q = Q_{0} e^{- \frac{t}{τ}} \Rightarrow \frac{Q_{2}}{8} = Q_{0} e^{- \frac{t_{2}}{τ}} \Rightarrow t_{2} 3 τ l n (2), a n d$

$U = \frac{Q^{2}}{2 C} = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow \frac{1}{2} (\frac{Q_{0}^{2}}{2 C}) = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow t_{1} = \frac{τ}{2} l n (2)$

$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{1}{6}$

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New answer posted

a year ago

Physics Dual Nature of Radiation and Matter

The electric field at a point associated with a light wave is given by

E = 200 $[s i n (6 * 1 0^{15}) t + s i n (9 * 1 0^{15}) t] V m^{- 1}$

Given : h = 4.14 * 10^-15 eVs

If the light fall on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

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Payal Gupta

Contributor-Level 10

The light wave contains two lights of different frequencies, so

$E_{1} = h υ_{1} = \frac{4.14 * 1 0^{- 15} * 6 * 1 0^{15}}{2 π} = 3.96 e V, a n d$

$E_{2} = h υ_{2} = \frac{4.14 * 1 0^{- 15} * 9 * 1 0^{15}}{2 π} = 5.92 e V$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

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