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$v=\frac{1}{\sqrt[]{\epsilon \mu }}=\frac{1}{\sqrt[]{2{\epsilon }_0.2{\mu }_0}}=\frac{1}{2\sqrt[]{{\epsilon }_0{\mu }_0}}=\frac{c}{2}=15\times 10^{7}m/s.$

$\frac{N{}_p}{N_s}=\frac{V_p}{V_s}\Rightarrow N_p=\frac{V_p}{V_s}\times N_s=\frac{220}{12}\times 24=440$

Assume the length of train be I and its acceleration be a.

$v^2=u^2+2al\Rightarrow al=\frac{v^2-u^2}{2}$

Velocity when middle point crosses the post,

$V_m=\sqrt[]{u^2+2a\frac{\mathcal{l}}{2}}.=\sqrt[]{u^2+\frac{v^2-u^2}{2}}=\sqrt[]{\frac{u^2+v^2}{2}}$

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So   $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$  

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2\times 3.14}{\sqrt[]{6.67\times 10^{-11}\times 6\times 10^{24}}}\left[{\left(8\times 10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7\times 10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

-> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$  

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21\times \frac{34}{7}=102$           

$v=\frac{p}{m}\Rightarrow v_p:v_d:v_{\alpha }=1:\frac{1}{2}:\frac{1}{4}=4:2:1$

$F_{mag}=qvB\Rightarrow F_P:F_d:F_{\alpha }=1\times 4:1\times 2:2\times 1=2:1:1$

In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here    $\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right|\begin{array}{cc}\hfill \begin{array}{l}=\left(42-22\right)-2\left(14+11\right)-3\left(-4-6\right)\\ \end{array}\hfill & \hfill =20-50+30=0\hfill \end{array}$

For infinite solution 

20a – 8b – 4c = 0 Þ 5a = 2b + c

[h] = ML²T^-1

[E] = ML²T^-2

[V] = ML²T^-2C^-1

[P] = MLT^-1

Class 12 physics chapter 1 Electric Charges and Fields is an important chapter for the CBSE Board exam. In unit 1 of the CBSE Board exam of class 12 Physics, there are questions from three chapters including - Electric Charges and Fields, Electrostatics, and Electrostatic Potential and Capacitance. This unit carries a total 16 marks. 

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

For first resonance,

$\mathcal{l}+0.3d=\frac{v}{4f}$

$\Rightarrow \mathcal{l}+0.3\times 6=\frac{336\times 100}{4\times 504}\Rightarrow \mathcal{l}=14.8cm$

Given $\frac{dy}{dx}=\frac{x{y}^2+y}{x}=y^2+\frac{y}{x}$  

OR   $\frac{dy}{dx}-\frac{y}{x}=y^{2}OR\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{x}.\frac{1}{y}=1.........\left(i\right)$          

->   Since curve intersect x + 2y = 4 at x = -2 then y = 3 so

From (ii) $\frac{-2}{3}=-2+cORc=2-\frac{2}{3}=\frac{4}{3}$  

put x = 3, then $\frac{3}{y}=\frac{-9}{2}+\frac{4}{3}=\frac{-19}{6}$  

$\Rightarrow y=\frac{-18}{19}$  

Magnetic field on the axis of a circular coil at distance x from centre, B = $\frac{{\mu }_{0}Ni{r}^2}{2{\left(r^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$\frac{{\left(r^2+{\left(0.2\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(r^2+{\left(0.05\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=8$ $\frac{r^2+{\left(0.2\right)}^2}{r^2+{\left(0.05\right)}^2}=4\Rightarrow r^2=0.01\Rightarrow r=0.1m.$  

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_1}{2^6}=\frac{N_2}{2^3}\Rightarrow \frac{N_1}{N_2}=8$

Given f (k) = $\{\begin{array}{l}k+1,kisodd\\ k,kiseven\end{array}$

  $?g:A\to A$           such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ × 1 = 10⁵

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$\Rightarrow 2=2\pi \sqrt[]{\frac{2}{g}}$

$\Rightarrow g=2{\pi }^{2}m/s^2$

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