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11 months ago

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

A wire $A$ , bent in the shape of an arc of a circle, carrying a current of $2 A$ and having radius $2 c m$ and another wire $B$ , also bent in the shape of arc of a circle, carrying a current of $3 A$ and having radius of $4 c m$ , are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and $B$ at the common centre $O$ is:

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alok kumar singh

Contributor-Level 10

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$ .

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11 months ago

Class 12th Thermodynamics

The total internal energy of two mole mono atomic ideal gas at temperature T = 300K will be __________J. (Given R = 8.31 J/mol.K)

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Payal Gupta

Contributor-Level 10

$U = n c_{v} T$

$\begin{array}{l} = 2 * \frac{3}{2} * R * 300 \\ = 7479 J \end{array}$

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Three

The number of matrices of order 3 * 3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is……….

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Vishal Baghel

Contributor-Level 10

$A = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}], a_{2}, b_{2}, c_{2} \in {0, 1}$

S = $a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3} + c_{1} + c_{2} + c_{3}$ is prime

$0 \leq s \leq 9$

Prime value = 2,3, 5, 7

$F o r S = 2, 1 + 1 + 0 + 0 + 0 + 0 + 0 + 0 + 0 \to \frac{9!}{2! 7!} = 36$

Total number of matrices

= 36 + 84 + 126 + 36 = 282

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11 months ago

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

A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2 Mg force. In this situation, tension in the string is $\frac{x}{5} M g$ for x_________. Neglect mass of the string and fiction of the block (bigger mass) with ice slab.

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Payal Gupta

Contributor-Level 10

Using newton's law:

2mg – T = 4ma - (1)

T – mg = ma - (2)

From (1) & (2) :

$m g = 5 m a \Rightarrow a = \frac{g}{5} = 2 m / s^{2}$

$T = m g + \frac{m g}{5} = \frac{6 m g}{5}$

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11 months ago

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

Particle $A$ of mass $m_{A} = \frac{m}{2}$ moving along the $x$ -axis with velocity $v_{0}$ collides elastically with another particle $B$ at rest having mass $m_{B} = \frac{m}{3}$ . If both particles move along the $x$ axis after the collision, the change $Δ λ$ in de-Broglie wavelength of particle $A$ , in terms of its de-Broglie wavelength $(λ_{0})$ before collision is:

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alok kumar singh

Contributor-Level 10

Sol. $\Rightarrow {\overset{?}{V}}_{1} = 2 \overset{?}{U_{c m}} - \overset{?}{U_{1}} \Rightarrow 2 [\frac{m / 2 V_{o}}{\frac{m}{2} + \frac{m}{3}}] - V_{o}$

$\Rightarrow \frac{6}{5} V_{o} - V_{o} \Rightarrow \frac{V_{0}}{5}$

$\Rightarrow λ_{o} = \frac{h c}{(\frac{m}{2} V_{o})} λ_{f} = \frac{h c}{(\frac{M}{2} \frac{V_{o}}{5})}$

$\Rightarrow Δ λ = \frac{8 h c}{{m V}_{o}}$

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11 months ago

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

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial

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Vishal Baghel

Contributor-Level 10

$T_{5} =^{n} C_{4} {(2^{\frac{1}{4}})}^{n - 4} . {({(\frac{1}{3})}^{\frac{1}{4}})}^{4} =^{n} C_{4} . 2^{\frac{n - 4}{4}} . \frac{1}{3}$

$T_{6} =^{9} C_{5} {(2^{\frac{1}{2}})}^{4} . {({(\frac{1}{3})}^{\frac{1}{4}})}^{5} =^{9} C_{5} . 2 . \frac{1}{3^{5 / 4}} = \frac{2}{3} . \frac{1}{3^{\frac{1}{4}}} . \frac{9 * 8 * 7 * 6}{4 * 3 * 2 * 1}$

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11 months ago

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

A man of 60kg is running on the road and suddenly jumps into stationary trolly car of mass 120kg. Then, the trolly car starts moving with velocity 2ms^-1. The velocity of the running man was _________ ms^-1, when he jumps into the car

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Payal Gupta

Contributor-Level 10

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒ v = 6m/s

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11 months ago

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

Le aa₁, a₂, a₃,…… be an A.P. If $\sum_{r = 1}^{\infty} \frac{a_{r}}{2^{r}} = 4,$ then 4a₂ is equal to……

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Vishal Baghel

Contributor-Level 10

$\sum_{r = 1}^{\infty} \frac{a + (r - 1) d}{2^{r}} = 4, 4 (a + d) = ?$

$\Rightarrow a \sum_{r = 1}^{\infty} {(\frac{1}{2})}^{r} + d \sum_{r = 1}^{\infty} (r - 1) {(\frac{1}{2})}^{r} = 4$

$= {(\frac{1}{2})}^{2} \sum_{k = 0}^{\infty} k {(\frac{1}{2})}^{k - 1}$

$= \frac{1}{4} . \frac{1}{{(1 - \frac{1}{2})}^{2}}$

= 1

a + d = 4

4 (a + d) = 16

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11 months ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Eleven

Let a line with direction ratios a, 4a, 7 be perpendicular to the lines with direction ratios 3, 1, 2b and b, a, 2. If the point of intersection of the line $\frac{x + 1}{a^{2} + b^{2}} = \frac{y - 2}{a^{2} - b^{2}} = \frac{z}{1}$ and the plane x – y + z = 0 is $(α, β, γ), t h e n α + β + γ$ is equal to………

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Vishal Baghel

Contributor-Level 10

(a, 4a, 7). (3, 1, 2b)= 0

3a + 4a – 14b = 0 a – 2b = 0 ……. (i)

(a, 4a, 7) . (b, a, 2) = 0

ab – 4a² + 14 = 0……… (ii)

(i) & (ii) =>2b² – 16b²+ 14 = 0

$\Rightarrow b^{2} = 1 \Rightarrow a = 2 b = \pm 2$

Plane : x – y + z = 0

P (4, 5, 1)

4 + 5 + 1 = 10

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11 months ago

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

Dimensional formula for thermal conductivity is (here $K$ denotes the temperature):

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alok kumar singh

Contributor-Level 10

$K = \frac{(\frac{Q}{r}) Δ x}{Δ A T}$

$\begin{array}{l} \Rightarrow & \frac{({M L}^{2} T^{- 2}) (L)}{(L^{2}) (θ) (T)} \\ \Rightarrow & M^{1} L^{1 -} T^{- 3} θ^{- 1} \end{array}$

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