Physics Test - 4

Given, diameter of circle, d = 50cm

= 50 x 10^-2 m

and frequency, f = 2Hz

The acceleration of particle in a uniform circular motion can be given as

a = ω² x

where, ω = angular frequency = 2πf

x = didance from centre = d/2

⇒ a = 4π² f² x d/2 …….(i)

Substituting given values in Eq. (i), we ge

a = 4π² x 4 x (50 x 10^-2 )/2 = 4π²

Given, focal length of convex lens, f₁ = f and focal length of concave lens, f₂ = -f.

The focal length of the combination, when two lenses are in contact is given 1/f = 1/f₁ + 1/f₂ ….(i)

Substituting given value in Eq. (i), we get

⇒ 1/f = 1/f - 1/f = 0 ⇒ f = ∞

Hence, the equivalent focal length of the combination is infinity.

To increase the range from I to nI a parallel resistance (shunt S) is connected across the galvanometer.

Given, work done. W = 12000J,

Initial frequency, f₁ = 10Hz and final frequency, f₂ = 20Hz

Angular velocity for rotational motion is given by ω = 2πf

∴ ω₁ = 2πf₁ = 2π x 10 = 20π rad/s and ω₂ = 2πf₂ = 2π x 20 = 40π rad/s

According to work-energy theorem, work done in rotation = change in rotational kinetic energy

.

The given circuit can be drawn a

Applying Kircbhoff’s voltage law (KVL) in loop ABCDA, we get

-(i₁ + i₂ )R +E₂ - i₂ r₂ = 0 ….(i)

Applying KVL in loop ABFEA we get -

- (i₁ + i₂ )R - i₁ r₁ + E₁ = 0 ….(ii)

Applying KVL in loop EFCDE, we get

-E₁ + i₁ r₁ + E₂ - i₂ r₂ = 0 ….(iii)

From the given options, only option (c) satisfies Eq..(ii), hence it is the correct equation.

The maximum stress produced in a rope is given by

σ_max = Force/Area = Mg/πr²

As the ift is accelerating with acceleration a, then

σ_max = M(g ± a)/πr²

⇒ r² = M(g ± a)/πS [Given , σ_max = S]

d² /4 = M(g ± a)/πS [∴ r = d/2]

As acceleration maximum i.e., g’ = g + a, so

Given, alternating voltage is

The voltage will be maximum, when

= T/6

Thus, the voltage will be maximum for the first time when t = T/6.

The balanced condition of the bridge is shown below in the first case

To get the balance point at 50 cm from the left end. the balanced condition becomes

As, in series combination resistances are added, so to get 9Ω at left gap the 3Ω resistance should be added in series with XΩ (6 + 3 = 9Ω).

Let two vectors, P and Q are represented by a graph as below

Here, is a vector in the direction of P.

Then, from the right angle triangle, we get

⇒O_p = Q cos θ

As given that, P is the unit vector along P, then

P = P/P ….(iii)

Putting the value of Ptrom Eq. (iii) to Eq. (ii), we get

Q_p = P ·Q

When a stream of air passes through the space between the baits, the pressure reduces between them as compared to the atmospheric pressure on either side. So, according to Bemoulli’s theorem, the pressure energy should be constant and is compensated by the movement of balls toward each other. So, the distance between the balls will decrease.

The critical angle is that angle on incidence in the denser medium for which the angle of refraction in rarer medium is 90º , i.e.

⇒ sin θ = μ_y /μ_x …(i)

where, μ = refractive index of the medium.

From Eqs, (i) and (iii), we get

Thus, the speed of light in medium y is

Using the given relation,

v = rω

If the perpendicular distance r from the axis of rotation is increased or decreased, then the value of linear velocity v accordingly increases or decreases. Thus, the value of angular velocity ω does not depend on r.

Given, Maximum wavelength,

λ_m = 289.8 nm = 289.8 x 10^-9 m

= 289.8 x 10^-10 m

Stefan's constant, σ = 567 x 10^-8 Wm^-2 K^-4

Wien's constant, b = 2898 μmK = 289.8 x 10^-6 mK

According to Wien’s displacement law. the maximum wavelength is given by

λ_m = b/T ⇒ T = b/λ_m ….(i)

Substituting given values in Eq. (i), we get

According to Stefan’s law, the energy radiated from a source is given by

E = σ AeT⁴ ...(ii)

where, A = area of source

e = emissivity (value between 0 to 1)

The intensity of radiations emitted is equal to energy radiated from a given surface area, i.e.,

l = E/A = σeT⁴ [from Eq.(iii)]

As e is very small, so

l = σeT⁴ …..(iv)

Substituting the value of T from Eq. (ii) in Eq. (iv), we get

l = σ(10⁴ )⁴ = 5.67 x 10^-8 x 10¹⁶

[∵ σ = 5.67 x 10^-8 (given)]

= 5.67 x 10⁸ Wm^-2

In frequency modulated waves, the only frequency varies with time, while the amplitude remains constant. This wave can be shown as

It is used in FM broadcasting, telemetry, and RADAR, etc.

The excess of pressure due to surface tension on a spherical liquid drop is given by p = 2T/R ….(i)

where. T = surface tension and

R = radius of the liquid drop