Physics

From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in __________ s.

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For first ball

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

$\Rightarrow - h = u * 6 - \frac{1}{2} * 10 * 6 * 6$

$\Rightarrow - h = 6 u - 180$

h = 180 - 6u -(1)

for second boy

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

$h = u * 1.5 + \frac{1}{2} * 10 * 1.5 * 1.5$

h = 1.5u + 11.25 -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

$u = \frac{168.75}{7.5} = 22.5$

h = 180 - 6u

= 180 - 6 * 22.5

45m

for third ball

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

$h = 0 * t + \frac{1}{2} * 10 t^{2}$

$\Rightarrow 45 = 5 t^{2}$

t² = 9

t = 3 sec

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6 months ago

Two identical charged particles each having a mass 10g and charge 2.0 * 10^-7 C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g = 10ms^-2]

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Contributor-Level 10

$μ m g = \frac{k q_{1} q_{2}}{L^{2}}$

$\Rightarrow 0.25 * 0.01 * 10 = \frac{9 * 1 0^{9} * 2 * 1 0^{- 7} * 2 * 1 0^{- 7}}{L^{2}}$

$= \frac{9 * 2^{2} * 1 0^{- 5 + 4}}{25 * 10}$

L = 12 cm

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6 months ago

When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $υ_{1} .$ When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $υ_{2} .$ If $υ_{2} = x υ_{1},$ the value of x will be __________.

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hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2} m v_{1}^{2} = h υ_{0}$

$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

$v_{1} = \sqrt[]{\frac{2 h υ_{0}}{m}}$ - (1)

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

v₂ = $\sqrt[]{\frac{8 h υ_{0}}{m}}$ - (2)

$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

v₂ = 2v₁

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6 months ago

A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

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Is minimum at the highest position of the circular path.

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The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be

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Contributor-Level 10

$T^{2} α R^{3}$

$\begin{array}{l} \Rightarrow \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{{R_{1}}^{3}}{{R_{2}}^{3}} \\ \Rightarrow \frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{R^{3}}{{(3 R)}^{3}} \\ T_{2} = 3 \sqrt[]{3} y e a r s \end{array}$

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6 months ago

Sodium light of wavelength 650nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maximum of diffraction pattern obtained in the two cases is _________* 10^-5 m.

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Contributor-Level 10

.For maxima = y = (2n + 1) $\frac{λ}{2} \frac{D}{a}$ $\frac{}{} \frac{}{}$

For 1st maxima for l1 wavelength (n = 1)

$_{} \frac{_{}}{} \frac{}{}$ $y_{1} = \frac{3 λ_{1}}{2} \frac{D}{a}$ - (1)

First maxima for l2 wavelength

$y_{2} = \frac{3}{2} λ_{2} \frac{D}{a}$ - (2)

$\Rightarrow y_{2} - y_{1} = \frac{3}{2} \frac{D}{a} [λ_{2} - λ_{1}]$

$= \frac{3}{2} * \frac{2 * 5 * 1 0^{- 9}}{0.5 * 1 0^{- 3}}$

$= \frac{3}{1 0^{- 3}} * 1 0^{- 8}$

$Δ y = 3 * 1 0^{- 5} m$

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6 months ago

Identify the pair of physical quantities that have same dimensions

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Contributor-Level 10

Velocity gradient = $\frac{d v}{d x} = \frac{L T^{- 1}}{L} = T^{- 1}$

Decay constant $(λ) = \frac{0.693}{T_{\frac{1}{2}}} = T^{- 1}$

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6 months ago

As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is $\frac{\sqrt[]{a}}{π} A .$ The value of a is __________.

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Contributor-Level 10

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{r m s} = \frac{l_{0}}{\sqrt[]{2}}$

$V_{r m s} = l_{r m s} z$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (X_{L})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (2 π * 50 * 200 * 1 0^{- 3})$

$\Rightarrow 220 = \frac{l_{0}}{\sqrt[]{2}} (20 π)$

$l_{0} = \frac{220 \sqrt[]{2}}{20 π}$

$l_{0} = \frac{11 \sqrt[]{2}}{π} = \frac{\sqrt[]{a}}{π}$

a = 121 * 2

a = 242

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6 months ago

A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kΩ. The voltage gain of the transistor will be ________.

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Contributor-Level 10

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 * 1 0^{- 3}}{10 * 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) * (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 * 1 0^{- 3}}{10 * 1 0^{- 6}} * \frac{5 * 1 0^{3}}{1 0^{3}}$

A_v = 750

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6 months ago

Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ________ cm.

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