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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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Payal Gupta

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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New answer posted

11 months ago

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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Payal Gupta

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

t = 7 * t_1/2

= 7 * 5 = 35 days

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New answer posted

11 months ago

Physics Laws of Motion

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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Payal Gupta

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

M * 0 = M₂vy + 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 question posted

11 months ago

Physics Laws of Motion

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New answer posted

11 months ago

Physics Motion in a Straight Line

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]

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Payal Gupta

Contributor-Level 10

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

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 + $\frac{1}{2} a t^{2}$

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

80 = 2u + 5 * 2 * 2

80 = 2u + 20

2u = 60

u = 30m/sec

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New answer posted

12 months ago

Physics Physics Oscillations

As per given figures, two spring constants k and 2k are connected to mass m. If period of oscillation in figure (a) is 3s, then the period of oscillation in figure (b) will be $\sqrt[]{x} s .$ The value of x is________.

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alok kumar singh

Contributor-Level 10

In figure a

$K_{e q} = \frac{k * 2 k}{3 k} = \frac{2 k}{3}$

$T = 2 Π \sqrt[]{\frac{3 M}{2 K}} = 3 s$

$\frac{I n f i g u r e b}{K_{e q} = 3 K}$

$T' = 2 Π \sqrt[]{\frac{M}{3 K}}$

$T' = \sqrt[]{2} \Rightarrow X = 2$

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New answer posted

12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Three

A potentiometer wire of length 300 cm is connected in series with a resistance 780 $Ω$ and a standard cell of emf 4V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is ________ $Ω$ .

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alok kumar singh

Contributor-Level 10

$I = \frac{4}{R + 780}$

Potential Difference across AB = l * R = $\frac{4 R}{R + 780}$

Potential Difference across AC = $(\frac{4 R}{R + 780}) * \frac{60}{300}$

$\Rightarrow \frac{4 R}{5 (R + 780)} = 2 * 1 0^{- 2} \Rightarrow R = 20 Ω$

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New answer posted

12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Six

In a coil of resistance $8 Ω,$ the magnetic flux due to an external magnetic field varies with time as $ϕ = \frac{2}{3} (9 - t^{2}) .$ The value of total heat produced in the coil, till the flux becomes zero, will be_______J.

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alok kumar singh

Contributor-Level 10

$? = \frac{2}{3} (9 - t^{2}) = 0$

t = 3 sec

$E m f = \frac{- d ?}{d t} = \frac{4 t}{3}$

$l = \frac{E m f}{R} = \frac{\frac{4}{3} t}{8} = \frac{t}{6}$

Heat = $\int l^{2} R d t = \int_{0}^{3} \frac{t^{2}}{36} * 8 d t = 2 J$

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New answer posted

12 months ago

Physics physics ncert exemplar solution class 12th chapter one

Three point charges of magnitude $5 μ C, 0.16 μ C, a n d 0.3 μ C$ are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3cm, BC = $3 \sqrt[]{2}$ and CA = 3 and point A is the right angle corner. Charge at point A experiences __________ N of electrostatic force due to the other two charges.

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alok kumar singh

Contributor-Level 10

$F_{1} = \frac{9 * 1 0^{9} * 5 * 1 0^{- 6} * 0.3 * 1 0^{- 6}}{9 * 1 0^{- 4}}$

= 15 N

$F_{2} = \frac{9 * 1 0^{9} * 5 * 1 0^{- 6} * 0.16 * 1 0^{- 6}}{9 * 1 0^{- 4}}$

= 8N

$\Rightarrow F_{n e t} = \sqrt[]{{(15)}^{2} + {(8)}^{2}}$

= 17 N

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New answer posted

12 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of $2 k Ω$ is connected in the collector branch of the circuit used. The input resistance of the transistor is $0.50 k Ω .$ The voltage gain of the transistor is________.

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alok kumar singh

Contributor-Level 10

Voltage gain = $\frac{Δ l_{C}}{Δ l_{B}} * \frac{R_{C}}{R_{B}}$

$\frac{5 * 1 0^{- 3}}{100 * 1 0^{- 6}} * \frac{2}{0.5}$

= 200

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