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11 months ago

Physics Class 11th

A ball of mass m is thrown vertically upwards. Another ball of mass 2 is thrown at an angle q with vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x} .$ The value of x is _______.

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Payal Gupta

Contributor-Level 10

$T_{1} = \frac{2 u}{g} T_{2} = \frac{2 V s i n θ}{g}$

$A s, T_{1} = T_{2} \Rightarrow \frac{2 u}{g} = \frac{2 V s i n θ}{g} \Rightarrow u = v s i n θ$ - (i)

$\frac{H_{1}}{H_{2}} = \frac{u^{2}}{2 g} * \frac{2 g}{v^{2} s i n^{2} θ} = {(\frac{u}{v s i n θ})}^{2} = 1$

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11 months ago

Physics Class 11th

A pulley of radius 1.5m is rotated about its axis by a force F = (12t – 3t²)N applied tangentially (while t is measured in seconds). If moment of inertia of the pulley about its axis of rotation is 4.5 kg m², the number of rotation made by the pulley before its direction of motion is reversed, will be $\frac{K}{π} .$ The value of K is ________.

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Payal Gupta

Contributor-Level 10

r = 1.5 m

F = 12t – 3t2 N

l = 4.5 kgm²

$α = \frac{T}{l} = \frac{\frac{2}{3} (12 t - 3 t^{2})}{4.5} = \frac{1}{3} (12 t - 3 t^{2})$

$\Rightarrow α = 4 t - t^{2}$

$\Rightarrow \frac{d w}{d t} = u t - t^{2} = \int_{0}^{0} d w = \int {(4 t - t^{2})}_{0}^{t} d t$

No. of rev = $\frac{36}{2 π} = \frac{18}{π} = \frac{k}{π} \Rightarrow k = 18$

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11 months ago

Physics Physics Mechanical Properties of Solids

A square aluminium (shear modulus is 25 * 10⁹ Nm^-2) slab of side 60 cm and thickness 15 cm is subjected to a shearing force (on its narrow face) of 18.0 * 10⁴ N. The lower edge is riveted to the floor. The displacement of the upper edge is ________ $μ m .$

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Payal Gupta

Contributor-Level 10

Shear modulus = 25 * 109 N/m2

$\frac{F}{A} = τ \frac{Δ x}{h} \Rightarrow Δ x = \frac{F h}{A τ} = \frac{18 * 1 0^{4} * 0.15}{3600 * 1 0^{- 4} * 25 * 1 0^{9}}$

$= \frac{15 * 1 0^{2}}{200 * 25 * 1 0^{5}}$

$= \frac{3 * 1 0^{2}}{5 * 200} * 1 0^{- 5}$

$= \frac{60}{200} * 1 0^{- 5} = 3 μ m$

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11 months ago

Physics Physics Oscillations

A mass 0.9kg, attached to a horizontal spring, executes SMH with an amplitude A₁. When this mass passes through its mean position, then a smaller mass of 124 g is placed over it both masses move together with amplitude A₂. If the ratio $\frac{A_{1}}{A_{2}} i s \frac{α}{α - 1},$ then the value of a will be______.

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Payal Gupta

Contributor-Level 10

$m_{1} = 900 g m, w_{1} = \sqrt[]{\frac{k}{. 9}}$

$m_{2} = 1024 g m, ω_{2} = \sqrt[]{\frac{k}{1.024}}$

$m_{1} v_{1} = m_{2} V_{2}$

$\Rightarrow m_{1} ω_{1} A_{1} = m_{2} ω_{2} A_{2}$

$\Rightarrow \frac{A_{1}}{A_{2}} = \sqrt[]{\frac{1024}{900}} = \frac{32}{30} = \frac{16}{15} = \frac{16}{16 - 1}$

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11 months ago

Physics Class 12th

A long cylindrical volume contains a uniformly distributed charge of density $ρ C m^{- 3} .$ The electric field inside the cylindrical volume at a distance $x = \frac{2 ε_{0}}{ρ} m$ from its axis is_________Vm^-1.

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Payal Gupta

Contributor-Level 10

$x = \frac{2 ε_{0}}{ρ}$

$E 2 π x l = \frac{ρ π x^{2} l}{ε_{0}}$

$E_{x} = \frac{ρ x^{2}}{2 x ε_{0}} = \frac{ρ x}{2 ε_{0}}$

$= \frac{ρ}{2 ε_{0}} * \frac{2 ε_{0}}{ρ} = 1 v / m$

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11 months ago

Physics Class 12th

A 1 m long copper wire carries a current of 1 A. If the cross section of the wire is 2.0mm² and the resistivity of copper is 1.7 * 10^-8 $Ω m,$ the force experienced by moving electron in the wire is _______* 10^-23N.

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Payal Gupta

Contributor-Level 10

i = 1 A $j = σ E$

A = 2mm²

$ρ = 1.7 * 1 0^{- 8} Ω m$ $F = e E = \frac{e j}{σ} = \frac{e i}{A σ} = \frac{e i ρ}{A}$

= $\frac{1.6 * 1 0^{- 19} * 1.7 * 1 0^{- 8}}{2 * 1 0^{- 6}}$

= 136 * 10^-23 N

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11 months ago

Physics Physics Alternating Current

To light, a 50 W, 100 V lamp is connected, in series with a capacitor of a capacitance $\frac{50}{π \sqrt[]{x}} μ F,$ with 200V, 50Hz AC source. The value of x will be________.

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Payal Gupta

Contributor-Level 10

$R = \frac{v_{2}}{P} = \frac{10000}{50} \Rightarrow R = 200 Ω$

$V = \sqrt[]{V_{C}^{2} + V_{R}^{2}}$

$\Rightarrow 200 = \sqrt[]{V_{C}^{2} + {(100)}^{2}}$

$x_{c} = \frac{1}{ω_{c}} = \frac{π \sqrt[]{x}}{100 π * 50} = \frac{\sqrt[]{x}}{5000 * 1 0^{- 6}} = \frac{\sqrt[]{x} * 1 0^{3}}{5}$

$= 200 \sqrt[]{x}$

$V_{C} = i x_{C}$

$\Rightarrow \sqrt[]{3} . \sqrt[]{1 + x} = 2 \sqrt[]{x}$

$\Rightarrow 3 (1 + x) = 4 x \Rightarrow x = 3$

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11 months ago

Physics Physics Wave Optics

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are p/2 and p/3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be________.

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Payal Gupta

Contributor-Level 10

$?_{A} = \frac{π}{2}, ?_{B} = \frac{π}{3}$

$I_{A} = I_{1} + I_{2} + 2 \sqrt[]{l_{1} l_{2}} c o s ? = l + 4 l + 2 \sqrt[]{l * 4 l} c o s \frac{π}{2} = 5 l$

$I_{B} = I + 4 I + 2 \sqrt[]{l . 4 l} c o s \frac{π}{3} = 5 l + 2 l = 7 l$

$Δ l = 7 l - 5 l = 2 l \Rightarrow x = 2$

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11 months ago

Physics Class 11th

The one division of main scale of Vernier calipers reads 1mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale, When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2cm and 6^th Vernier division coincides with a main scale division. The diameter of the bob will be ________* 10^-2 cm.

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Payal Gupta

Contributor-Level 10

1 msD = 1mm

10 vsD = 9msD

1vsD = 0.9 MsD

L.C. = 1MSD – 1VSD = 1 – 0.9 = 0.1 mm

Zero error = 4LC = 0.4 mm

Reading = MSR + VSR + correction

= 4.1 cm + 6 * .01 cm + (-0.04 cm) = (4.1 + 0.06 – 0.04) cm

= 4.12 cm = 412 * 10^-² cm

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11 months ago

Physics Class 12th

In a meter bridge experiment, for measuring unknown resistance 'S', the null point is obtained at a distance 30cm from the left side as shown at point D. If R is 5.6 $k Ω$ , then the value of unknown resistance 'S' will be ___________ $Ω .$

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Payal Gupta

Contributor-Level 10

$\frac{30}{70} = \frac{S}{5.6 k Ω} \Rightarrow S = \frac{3}{7} * 5.6 k Ω = 3 * 0.8 k Ω$

= 2.4k $Ω$ = 2400 $Ω$

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