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2 months ago

Physics Waves

In an experiment to determine the velocity of sound in air at room temperature using a resonance tube, the first resonance is observed when the air column has a length of 20.0 cm for a tuning form of frequency 400 Hz is used. The velocity of the sound at room temperature is 336 ms^-1. The third resonance is observed when the air column has a length of ____________cm.

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Vishal Baghel

Contributor-Level 10

According to Concept of resonance tube, we can write

$\frac{λ}{4} + e = l_{1} a n d \frac{3 λ}{4} + e = l_{2} \Rightarrow \frac{λ}{2} = l_{2} - l_{1} \Rightarrow \frac{V}{2 ν} = l_{2} - l_{1}$

$\Rightarrow l_{2} = \frac{v}{2 ν} + l_{1} = \frac{336}{400} + 0.20 = 0.84 + 0.20 = 1.04 m = 104 c m$

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2 months ago

Physics Mechanical Properties of Fluids

A small spherical ball of radius 0.1mm and density 10⁴kgm^-3 falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity inside water, then the value of h will be_________ m.

(Given g = 10 ms^-2, viscosity of water = 1.0 * 10^-5 N-sm^-2).

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Vishal Baghel

Contributor-Level 10

Since velocity does not change, so acceleration will be zero.

mg = FB + Fv $\Rightarrow \frac{4 π}{3} r^{3} ρ g = \frac{4 π}{3} r^{3} σ g + 6 π η r v$

$\Rightarrow v = \frac{2 r^{2} (ρ - σ) g}{9 η} = \frac{2 * 0.1 * 0.1 * 1 0^{- 6} * (1 0^{4} - 1 0^{3}) * 10}{9 * 1.0 * 1 0^{- 5}}$

$\Rightarrow h = \frac{400}{2 g} = 20 m$

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2 months ago

Physics Units and Measurement

The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of (-0.05) cm. White measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The correct diameter will be__________ * 10^-2 cm.

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Vishal Baghel

Contributor-Level 10

Least count of Vernier = 0.1mm

Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

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2 months ago

Physics Physics Motion in Plane

A person can throw a ball upto a maximum range of 100m. How high above the ground he can the same ball?

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$R = \frac{u^{2} s i n 2 θ}{g} \Rightarrow u^{2} = g R_{m a x}, w h e r e θ = 45 °$

$\Rightarrow H_{m a x} = \frac{u^{2}}{2 g} = \frac{g R_{m a x}}{2 g} = \frac{R_{m a x}}{2} = \frac{100}{2} = 50 m$

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2 months ago

Physics Class 12th

The combination of two identical cells, whether connected in series or parallel combination provides the same current through an external resistance of $2 Ω .$ The value of internal resistance of each cell is

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Vishal Baghel

Contributor-Level 10

According to question, we can write

$\frac{2 V}{2 + 2 r} = \frac{V}{2 + \frac{r}{2}}$

$\Rightarrow 2 + 2 r = 4 + r \Rightarrow r = 2 Ω$

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2 months ago

Physics Class 12th

If the electric potential at any point (x, y, z)m in space is given by V = 3x² volt. The electric field at the point (1, 0, 3)m will be:

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Vishal Baghel

Contributor-Level 10

According to relation between field and potential, we can write

^{$\vec{E} = - \frac{d V}{d x} (\hat{i}) = - \frac{d (3 x^{2})}{d x} (\hat{i}) = - 6 x (\hat{i})$}

^{${\vec{E}}_{(1, 0, 3)} = - 6 (\hat{i}) N / C$}

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2 months ago

Physics Class 11th

A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box.

The value of 'a' will be

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Vishal Baghel

Contributor-Level 10

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

ma = mg – N = mg - $\frac{m g}{4} = \frac{3 m g}{4}$

$\Rightarrow a = \frac{3 g}{4}$

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2 months ago

Physics Work, Energy and Power

In the given figure, the block of mass m is dropped from the point 'A'. The expression for kinetic energy of block when it reaches point 'B' is

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Vishal Baghel

Contributor-Level 10

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

=>K_f – K_in = U_in - U_f

=>K - = mgy – mg (y – y₀) = mgy₀

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2 months ago

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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Vishal Baghel

Contributor-Level 10

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

4a = 4g – T . (1), and

40a = T - f_k = 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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2 months ago

Physics Physics Electric Charge and Field

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

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

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Raj Pandey

Contributor-Level 9

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

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