Physics

$tan\theta =\frac{v_y}{v_x}=\frac{u_y-gt}{u_x}$

$\Rightarrow tan\theta =\left|\frac{8-10}{4}\right|=\left|-\frac{1}{2}\right|$

(1)      2.3056 -> 4 decimals

10.138 -> 3 decimals

-7.4671 -> 4 decimals

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4.9765

Answer should have 3 decimals.

(2)  2.38 * 1.0 = 2.38 -> answer should have 2 significant digits

(3)   $\frac{8.05}{3.1}=2.59\approx 2.6$  answer should have 2SD

(4)  1.11 - 0.1 = 1.01 $\approx$  1.0 but is an intermediate step so we keep 1 digit extra. 1.01 * 9.0 = 9.09 $\approx$  9.1 -> both 1.01 and 9.0 have 2SD

Kindly consider the following figure

Kindly consider the following figure

Kindly consider the following figure

$\overrightarrow{L}=\overrightarrow{r}*m\overrightarrow{v}$

= $\left(3\widehat{i}-\widehat{j}\right)*\left(3\widehat{j}+\widehat{k}\right)$

$=9\widehat{k}+3\left(-\widehat{j}\right)-\widehat{i}$

$=-\widehat{i}-3\widehat{j}+9\widehat{k}$

$\left|\overrightarrow{L}\right|=\sqrt[]{1+9+81}=\sqrt[]{91}$

m = 0.3g density,   ${\rho }_{\left(ball\right)}=8g/cc$                   $\therefore V=\frac{m}{\rho }=\frac{0.3}{8}cc$                     

${\rho }_{glycerene}=1.3g/cc=1.3*10^{3}kg/m^3$

F_v + F_B = mg.

Þ F_v = mg – F_B = .3 * 10^-3 * 10 – 1.3 *  $\frac{0.3}{8}*10^{-6}*10*10^3$

  $=3*10^{-3}-\frac{.39}{8}*10^{-2}=3*10^{-3}-.5*10^{-3}$

= 2.5 * 10^-3N

we know, = $\frac{\lambda }{d}$ $\frac{}{}$ $\mu =\frac{C}{V}=\frac{C}{\upsilon \lambda }$ $\frac{}{}\frac{}{}$

${}_{}\frac{}{{}_{}{}_{}}$  ${\lambda }_1=\frac{C}{{\mu }_1{\upsilon }_1}$ [with change in medium frequency remain constant]

${\lambda }_2=\frac{C}{{\mu }_2{\upsilon }_2}$

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{{\mu }_1}{{\mu }_2}=\frac{5}{7}$ ${\lambda }_2=\frac{5}{7}{\lambda }_1$

${\theta }_1=\frac{{\lambda }_1}{d_1}$

2= $\frac{{\lambda }_2}{d_2}$

$\frac{{\theta }_2}{{\theta }_1}=\frac{{\lambda }_2}{{\lambda }_1}[?d_1=d_2]$

2 $=\frac{5}{7}*{\theta }_1$

${\theta }_2=\frac{5}{7}*0.35$

$\frac{B_0^2}{2{\mu }_0}C$

$B_0^2=\frac{I2{\mu }_0}{C}$

$B_0^2=\frac{0.22*2*4*10^{-7}}{9*108}$

B₀ = 42.9 * 10^-9

42.9 Ans

Least  count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.