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11 months ago

Physics Alternating Current Overview

For a series LCR circuit, I vs curve is shown:

Choose the most appropriate answer from the options given below:

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alok kumar singh

Contributor-Level 10

At w_r impedance of the circuit is equal to the resistance of the circuit.

Left of w_r, circuit is mainly capacitive

Right of w_r, the circuit is mainly Inductive.

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New answer posted

11 months ago

Physics Physics Moving Charges and Magnetism

A charge particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charge particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charge particle:

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alok kumar singh

Contributor-Level 10

$\sqrt[]{k \cdot E}$

$\frac{r_{f}}{r_{0}} = \frac{\sqrt[]{k . E_{f}}}{k . E_{i}}$

$\frac{r_{f}}{r_{0}} = \frac{\sqrt[]{4 k . E_{i}}}{\sqrt[]{k . E_{i}}}$

$\frac{r_{f}}{r_{0}} = \frac{2}{1} = 2 : 1$

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11 months ago

Physics Work, Energy and Power

A particle of mass 500 gm is moving in a straight line with velocity = The work done by the net force during its displacement from x = 0 to x = 4m is:

(Take b = 0.25m^-3/2s^-1).

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alok kumar singh

Contributor-Level 10

$ω = Δ k . E = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$= \frac{1}{2} m {[b * 4^{5 / 2}]}^{2} - \frac{1}{2} m {[0]}^{2}$

$= \frac{1}{2} * 0.5 [0.2 5^{2} * 4^{5}] = 2^{4} = 16 J$

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New answer posted

11 months ago

Physics Dual Nature of Radiation and Matter

Given below are two statements : one is labelled as Assertion A and other is labelled as Reason R.

Assertion A : The photoelectric effect does not takes place, if the energy of the incident radiation is less than the work function of a metal.

Reason R : Kinetic energy of the photoelectrons is zero, if the energy of the incident radiation is equal to the work function of a metal.

In the light of the above statements, chose the most appropriate answer from t

...more

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alok kumar singh

Contributor-Level 10

To free the electron from metal surface minimum energy required, is equal to the work function of that metal.

So Assertion A, is correct

have = w₀ + K.E_max

If have = w₀

Þ K.E_max = 0

Hence reas

...more

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11 months ago

Physics Physics Mechanical Properties of Solids

A wire of length L is hanging from a fixed support. The length changes to L₁ andL₂when masses 1kg and 2kg are suspended respectively from its free end. Then the value of L is equal to:

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alok kumar singh

Contributor-Level 10

Spring constant of wire (k) = $\frac{A E}{L}$

Case 1

$k [Δ] = 1 g$

$k [L_{1} - L] = 1 g$ - (1)

Case 2 $k [L_{2} - L] = 2 g$ - (2)

$(1) \div (2)$

$\frac{L_{1} - L}{L_{2} - L} = \frac{1}{2}$

L = 2L₁ – L₂

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New answer posted

11 months ago

Physics Thermodynamics

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)

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alok kumar singh

Contributor-Level 10

44.4 litres of He $\frac{44.8}{22.4} m o l e s$

$η = 2$ moles of He

Q = $η C_{v} Δ T$ (for fixed capacity (v = constant)

= $2 * \frac{R}{γ - 1} Δ T$

= $\frac{2 * R}{\frac{5}{3} - 1} * 20$

= $= 2 * \frac{3 R}{2} * 20$

Q = 60R = 60 * 8.3

Q = 498 J

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New answer posted

11 months ago

Physics System of Particles and Rotational Motion

A spherical of 1 kg mass and radius R is rolling with angular speed $ω$ on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is $\frac{a}{3} R^{2} ω .$ The value of a will be:

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alok kumar singh

Contributor-Level 10

L₀ = RMV_cm + I_cm $ω$

= RM $(R ω) + \frac{2}{3} M R^{2} ω$

L₀ = $\frac{5}{3} M R^{2} ω$

$\frac{5}{3} * 1 R^{2} ω = \frac{5}{3} R^{2} ω = \frac{a}{3} R^{2} ω$ =

a = 5

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11 months ago

Physics Physics Nuclei

The activity of a radioactive material is 2.56 * 10^-³ Ci. If the half life of the material is 5 days, after how many days the activity will become 2 * 10^-⁵ Ci?

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alok kumar singh

Contributor-Level 10

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

&nbs

...more

A₀ = 2.56 * 10^-3C_i

A = 2 * 10^-15C_i

A = ^{$\frac{A_{0}}{2^{n}}$}

$\Rightarrow 2 * 1 0^{- 5} = \frac{2.56 * 1 0^{- 3}}{2^{n}}$

2ⁿ = 128

n = 7

t = 7 * t_1/2

= 7 * 5 = 35 days

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New answer posted

11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

A body of mass M at rest explodes into pieces, in the ratio of masses 1 : 1: 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms^-¹ and 40 ms-¹ respectively. The velocity of the third piece will be:

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alok kumar singh

Contributor-Level 10

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

&nb

...more

Let 'v' be the velocity of third piece

com along x-axis

M * 0 = M₁v_x + M₃v'_x

$\Rightarrow 0 = M_{1} * 30 + M_{3} v' x$

$v_{x} 1 = \frac{- M_{1}}{M_{3}} * 30$

$\frac{- 1}{2} * 30$

$v_{x}^{1} = - 15 m / s e c$

Com along y-axis

M * 0 = M₂v_y + M₃v'_y

v_y = $\frac{- M_{2}}{M_{3}} v_{y}$

v_y = $\frac{- 1}{2} * 40$

v_y = -20m/sec

^{$\vec{v} = - 15 \hat{i} - 20 \hat{j}$}

v = $\sqrt[]{1 5^{2} + 2 0^{2}} = \sqrt[]{225 + 400} = 25 m / s e c$

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New answer posted

11 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

Two balls A and B placed at the top of 180m tall tower. Ball A is from the top at t = 0s. Ball B is thrown vertically down with an initial velocity 'u' at t = 2s. After a certain time, both balls meet 100m above the ground. Find the value of 'u' in ms^-¹.

[use g = 10 ms^-²] :

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

...more

Let 't' be the time taken by B to meat A.

for A, u = 0

t = [2 + t]

h = 80m

a = 10 m/sec²

h = ut + $\frac{1}{2} a t^{2}$

80 = 0 * t + $\frac{1}{2} (10) {(2 + t)}^{2}$

16 = (2 + t)²

t + 2 = 4

t = 2sec

Now for B

h = ut + $\frac{1}{2} a t^{2}$

$? 80 = u * 2 + \frac{1}{2} * 10 t^{2}$

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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