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8 months ago

Physics Physics Electrostatic Potential and Capacitance

A composite parallel plate capacitor is made up to two different dielectric materials with different thickness (t₁ and t₂) as shown in figure. The two different dielectric materials are separated by a conducting foil F. The voltage of the conducing foil is ___________V.

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Payal Gupta

Contributor-Level 10

Capacitance of each capacitor

$C_{1} = \frac{A 3 ε_{0}}{\frac{1}{2}} = 6 A ε_{0}$

$C_{2} = A 4 ε_{0} = 4 A ε_{0}$

Equivalent capacitance

$Δ v_{2} = \frac{240 A ε_{0}}{4 A ε_{0}} = 60 v$

$v_{f o i l} = 60 v$

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New answer posted

8 months ago

Physics Physics Wave Optics

When a car is approaching the observer, the frequency of horn is 100Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be $\frac{x}{3}$ Ha where x = _________.

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Payal Gupta

Contributor-Level 10

$f_{1} = 100 = f o (\frac{C}{C - V_{S}})$

C = Speed of sound

Vs = Speed of source

f2 = 50 = fo = $(\frac{C}{C + V_{s}})$

$\frac{f_{1}}{f_{2}} = 2 = \frac{C + V_{S}}{C - V_{S}}$

$f o = \frac{200}{3} = \frac{x}{3}$

x = 200

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New answer posted

8 months ago

Physics Units and Measurement

In an experiment to determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young's modulus $(Δ Y)$ is found to be x * 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

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Payal Gupta

Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 103 m

$\frac{F}{A} = y \frac{Δ L}{L}$

$y = \frac{F L}{A Δ L} = \frac{(m g) * 1}{(\frac{π d^{2}}{4}) * 0.4 * 1 0^{- 3}}$

$Δ y = 0.1 * 0.199 * 1 0^{12} = 1.99 * 1 0^{10}$

= 1.99

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New answer posted

8 months ago

Physics Physics System of Particles and Rotational Motion

A disc of mass 1kg and radius R is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of disc. A body of same mass as that of disc is fixed at the highest point of the disc. Now the system is released, when the body comes to the lowest position, its angular speed will be $\sqrt[]{\frac{x}{3 R} r a d s^{- 1}}$ where x = __________.

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Payal Gupta

Contributor-Level 10

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2} (\frac{1}{2} m R^{2} + m R^{2}) ω^{2}$

$ω = \sqrt[]{\frac{8 g}{3 R}} = 4 \sqrt[]{\frac{g}{2 * 3 R}}$

$x = \frac{g}{2} = 5$

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New answer posted

8 months ago

Physics Class 11th

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

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Payal Gupta

Contributor-Level 10

y = x5 (1 – x) = x tan θ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y – component of initial velocity

= u sin θ

= $= \sqrt[]{26} * \frac{5}{\sqrt[]{26}}$

= 5 m/s

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New answer posted

8 months ago

Physics Units and Measurement

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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Payal Gupta

Contributor-Level 10

Least count = $\frac{0.5}{50} m m$ = 0.01 mm

Diameter, d = 1.5 + 7 * 0.01

= 1.57

$∴$ Surface Area = (2r) $l$

= 3.4 cm²

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New answer posted

8 months ago

Physics Class 11th

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

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Payal Gupta

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow$ $\frac{u_{1}}{u_{2}}$ $=2$

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New answer posted

8 months ago

Physics Class 11th

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

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Payal Gupta

Contributor-Level 10

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

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New answer posted

8 months ago

Physics Work, Energy and Power

As per the given figure, two blocks each of mass 250g are connected to a spring of spring constant 2Nm^?1. If both are given velocity in opposite directions, then maximum elongation of the spring is

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Payal Gupta

Contributor-Level 10

Let m = 250 g = 0.25 kg

By Reduced mass method

$m_{r} = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = \frac{m m}{m + m} = \frac{m}{2}$

By wet

$w_{S P} = Δ K . E .$

$- \frac{1}{2} k x^{2} = 0 - \frac{1}{2} (\frac{m}{2}) {(2 v)}^{2}$

$\frac{2}{2} x^{2} = 0.25 v^{2}$

$x^{2} = 0.25 v^{2}$

$x = \frac{v}{2}$

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New answer posted

8 months ago

Physics Physics Gravitation

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth will, be

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Payal Gupta

Contributor-Level 10

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 * \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} * 100 = 1 %$

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