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6 months ago

Class 12th Motion in a Plane

If $\bar{A} = (2 \hat{i} + 3 \hat{j} - \hat{k}) m a n d \vec{B} = (\hat{i} + 2 \hat{j} + 2 \hat{k}) m .$ The magnitude of component of vector $\vec{A}$ along vector $\vec{B}$ will be________m.

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alok kumar singh

Contributor-Level 10

Component of $\vec{A}$ along $\vec{B}$ = $\frac{\vec{A} . \vec{B}}{| \vec{B} |}$

= 2

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6 months ago

Class 12th physics ncert exemplar solutions class 12th chapter two

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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Vishal Baghel

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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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The area of cross section of the rope used to lift a load by a crane is 2.5 * 10^-4m². The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be:

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alok kumar singh

Contributor-Level 10

Breading stress = $\frac{W e i g h t}{A r e a}$

$\Rightarrow \frac{10}{2.5 * 10.4} = \frac{25}{A}$

$\Rightarrow A = 6.25 * 1 0^{- 4} m^{2}$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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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Vishal Baghel

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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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

Mass numbers of two nuclei are in the ratio of 4 : 3. Their nuclear densities will be in the ratio of

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alok kumar singh

Contributor-Level 10

$R = R ? A^{1 / 3}$

$δ = \frac{m a s s}{v o l u m e} = \frac{m A}{\frac{4}{3} π R^{3}} = \frac{m A}{4 / 3 π R_{0}^{3} A}$

$\Rightarrow$ Density is independent of mass number

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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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Vishal Baghel

Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 10^-3 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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6 months ago

Class 12th General Principles & Processes of Isolating Elemen

Match List – I with List – II.

List – I List – II

(A) Siderite (i) Cu

(B) Calamine (ii) Ca

...more

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Payal Gupta

Contributor-Level 10

Siderite – FeCO₃ (ore of iron)

Calamine – ZnCO₃ (ore of zinc)

Malachite – CuCO₃.Cu (OH)₂ (ore of copper)

Cryolite – Na₃AlF₆ (ore of aluminium)

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Five

Two masses M₁ and M₂are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M₂is twice that of M₁. The acceleration of the system is a₁. When the mass M₂ is thrice that of M₁ the acceleration of the system is a₂. The ratio $\frac{a_{1}}{a_{2}}$ will be:

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alok kumar singh

Contributor-Level 10

$a_{1} = \frac{m_{2} g - m_{1} g}{m_{1} + m_{2}} = \frac{2 m g - m g}{3 m} = \frac{g}{3}$

$a_{2} = \frac{3 m g - m g}{4 m} = \frac{g}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{2}{3}$

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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Seven

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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Vishal Baghel

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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6 months ago

Class 12th Physics NCERT Exemplar Solutions Class 12th Chapter Four

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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Vishal Baghel

Contributor-Level 10

y = x5 (1 - x) = x tanθ $(1 - \frac{x}{R})$

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 sinθ

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

= 5 m/s

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