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9 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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9 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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9 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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9 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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9 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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9 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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9 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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9 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

In the circuit shown below, maximum zener diode current will be _________mA.

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Vishal Baghel

Contributor-Level 10

$l = \frac{120 - 60}{4000} = 0.015 A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA

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9 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A uniformly heavy rod of mass 20kg, cross sectional area 0.4 m² an length 20m is hanging from a fixe support. Neglecting the lateral contraction. The elongation in the rod due to its own weight is x * 10^-9m. The value of x i________.

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alok kumar singh

Contributor-Level 10

$\int_{0}^{Δ l} d y = \int_{0}^{l} (\frac{m g}{l}) * \frac{d x}{A Y}$

$\Rightarrow Δ l = \frac{m g l}{2 A Y} = 25 * 1 0^{- 9} m$

$\Rightarrow x = 25.00$

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9 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ${}_{1}^{2}X +_{1}^{2} X = \frac{4}{2} Y$

The binding energies per nucleon for are 1.1 Me V and 7.6

Me V respectively. The energy released in this process is________ Me V.

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alok kumar singh

Contributor-Level 10

Energy released = Binding energy of product – Binding energy of reacts :

$Δ E = 7.6 * 4 - (1.1 * 4)$

= 26 MeV

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