Physics NCERT Exemplar Solutions Class 12th Chapter Three

 $h=\frac{u^2}{2g}\Rightarrow u=\sqrt[]{2gh}$

$\frac{h}{3}=\sqrt[]{2gh}t-5t^2$

$5t^2-\sqrt[]{20h}t+\frac{h}{3}=0$

$t=\frac{\sqrt[]{20h}\pm \sqrt[]{20h-4*5*\frac{h}{3}}}{10}$

$\therefore \frac{t_1}{t_2}=\frac{\sqrt[]{20h}-\sqrt[]{\frac{40}{3}h}}{\sqrt[]{20h}+\sqrt[]{\frac{40}{3}h}}=\frac{1-\sqrt[]{\frac{2}{3}}}{1+\sqrt[]{\frac{2}{3}}}=\frac{\sqrt[]{3}-\sqrt[]{2}}{\sqrt[]{3}+\sqrt[]{2}}$

 $t=\sqrt[]{x}+4$

then $\frac{dt}{dx}=\frac{1}{2\sqrt[]{x}}\therefore {\left(\frac{dx}{dt}\right)}_{t=4}={\left(2\sqrt[]{x}\right)}_{t=4}={\left[2\left(t-4\right)\right]}_{att=4}$

= 0

Sol. $\stackrel{?}{\tau }=\stackrel{?}{\mu }*\stackrel{?}{B}$

$\begin{array}{ll}\Rightarrow & 0.018=\mu \left(0.06\right)\left(\mathrm{s}\mathrm{i}\mathrm{n}?{30}^{?}\right)\\ \Rightarrow & \mu =0.6\\ \Rightarrow & \text{Work}=U_f-U_i\\ \Rightarrow & 2\mu B\\ \Rightarrow & 7.2*{10}^{-2}\mathrm{J}.\end{array}$

Total time (T) = 4 sec

Given u = 0 m/sec

a = g = 9.8 m/sec²

h = 4.9 m, t =?

$\Rightarrow h=ut+\frac{1}{2}a{t}^2$                

$4.9=0.t+\frac{1}{2}*9.8t^2$                

t = 1 sec

'v' be the velocity with which ball hits the water v = u + at

= 0 + 9.8 * 1 = 9.8 m/sec

Time taken to reach the bottom of the lake from surface of the lake

= 4 – 1 = 3 sec

v = 9.8 m/sec

$H=ut+\frac{1}{2}g{t}^2=9.8*3+\frac{1}{2}*9.8*9$                

29.4 + 4.9 * 9 = 29.4 + 44.1

H = 73.5 m

NCERT Exemplar goes beyond the NCERT textbook as it contains questions based on the textbook concepts. While practicing the short answer type questions, long answer type questions, and multiple choice questions, students learn how to use these concepts for solving problems. It will help students improve their analytical thinking and problem-solving skills and develop a deeper understanding of concepts. It prepares students for CBSE Board exams and competitive exams like NEET and JEE.

While applying Kirchhoff's Current Law (KCL) at junctions, the total current leaving is equal to the total current entering. For Kirchhoff's Voltage Law (KVL) in loops, the total of the voltage gains and drops around any closed loop is zero. To avoid mistakes, students must assign a direction to currents which can be positive for gains and negative for drops, and be consistent with the signs while solving the equations.

When no current flows in the circuit, the total potential difference provided by a source is known as the Electromagnetic force (EMF). On the other hand, when current flows in the circuit, the actual voltage across the terminals of the cell is called the terminal voltage. The formula that depicts the relation is:

Where r is the internal resistance and I is the current.

Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R1}{R2}=\frac{R3}{R4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- according to the relation $\rho$ = $\frac{m}{ne2t}$ also T $\propto \frac{1}{t}$