Question

Plz answer this. The best answer will be marked branliest

Answer 1

Fig :-

$\setlength{\unitlength}{1 cm}\begin{picture}(12,6)\put(7.5,7.5){\circle{4}}\put(7.5,7.5){\circle{1}}\put(7.5,7.5){\line(1,0){1}}\put(7.5,7.5){\line(-1,0){1}}\put(7.5,7.5){\line(0,1){1}}\put(7.5,7.5){\line(0,-1){1}}\put(7.5,7.5){\circle*{0.1}}\put(7.5,7.5){\line(3,4){0.45}}\put(7.5,7.5){\line(-3,-4){0.45}}\put(7.7,7.52){ \theta }\put(7.15,7.22){ \theta }\put(7.1,6.6){ \rho_1 }\put(8.2,7.1){ \rho_2 }\put(7.75,7.1){\line(1,0){0.32}}\end{picture}$

❏ Solution:-

➩ generalizing the figure ,

$\setlength{\unitlength}{1 cm}\begin{picture}(12,6)\put(7.5,7.5){\circle{4}}\put(7.5,7.5){\circle{1}}\put(7.5,7.5){\line(1,0){1}}\put(7.5,7.5){\line(-1,0){1}}\put(7.5,7.5){\line(0,1){1}}\put(7.5,7.5){\line(0,-1){1}}\put(7.5,7.5){\circle*{0.1}}\put(7.5,7.5){\line(4,3){0.56}}\put(7.5,7.5){\line(-4,-3){0.56}}\put(7.72,7.5){ \theta }\put(7.11,7.26){ \theta }\put(7.5,6.8){\line(1,0){2.7}}\put(7.5,6.8){\line(-1,0){2.7}}\put(8.05,8.09){\line(1,0){0.3}}\put(8.5,8.09){\line(1,0){0.3}}\put(8.95,8.09){\line(1,0){0.3}}\put(9.4,8.09){\line(1,0){0.3}}\put(8.25,7.1){\line(1,0){0.2}}\put(8.65,7.1){\line(1,0){0.2}}\put(9.05,7.1){\line(1,0){0.2}}\put(9.5,7.1){\line(1,0){0.2}}\put(9.95,7.1){\line(1,0){0.2}}\put(9.8,7.6){\vector(0,1){0.5}}\put(9.8,7.6){\vector(0,-1){0.5}}\put(6.8,7.07){\line(-1,0){0.2}}\put(6.5,7.07){\line(-1,0){0.2}}\put(6.2,7.07){\line(-1,0){0.2}}\put(5.9,7.07){\line(-1,0){0.2}}\put(5.6,7.07){\line(-1,0){0.2}}\put(5.3,7.07){\line(-1,0){0.2}}\put(5,7.07){\line(-1,0){0.2}}\put(9.9,7.6){ h }\put(4.4,6.84){ h_1 }\put(10.3,6.84){ h_2 }\put(7.1,6.6){ \rho_1 }\put(8.2,7.25){ \rho_2 }\put(7.75,7.1){\line(1,0){0.32}}\end{picture}$

Now , we know that

$pressure = depth\times density\times gravitational \:acceleration$

Therefore applying the theorem ,

$\blacksquare\:\:\sf{\ \ {P_1=h_1\rho_1g}}$

And

$\blacksquare\:\:\sf{\ \ {P_2=h_2\rho_1g+h\rho_2g}}$

As the liquids are in rest so , $P_1=P_2$ ,

$\blacksquare\:\:\sf{\ \ {P_1=P_2}}$

$\implies\:\:\sf{\ \ {h_1\rho_1g=h_2\rho_1g+h\rho_2g}}$

$\implies\:\:\sf{\ \ {h_1\rho_1=h_2\rho_1+h\rho_2}}$ --------------------(i)

Now , generalized the figure more , we get ,

$\setlength{\unitlength}{1 cm}\begin{picture}(12,6)\put(7.5,7.5){\line(1,0){2}}\put(7.5,7.5){\line(-1,0){2}}\put(7.5,7.5){\line(0,1){2}}\put(7.5,7.5){\line(0,-1){2}}\put(7.5,7.5){\circle*{0.1}}\put(7.5,7.5){\line(4,3){1}}\put(7.5,7.5){\line(-4,-3){1}}\put(7.5,7.5){\line(4,-3){1}}\put(7.8,7.5){ \theta }\put(7.0,7.23){ \theta }\put(8.48,7.5){\line(0,1){0.74}}\put(8.48,7.5){\line(0,-1){0.74}}\put(6.52,7.5){\line(0,-1){0.74}}\put(7.5,6.4){\line(1,0){2.8}}\put(7.5,6.4){\line(-1,0){2.8}}\put(7.52,7.1){ \theta }\put(8.28,6.9){ \theta }\put(7.75,7.9){ R }\put(8.6,7.7){ R\sin\theta }\put(8.6,7.0){ R\cos\theta }\put(7.15,6.85){ R }\put(6.4,6.75){\line(-1,0){0.2}}\put(6.1,6.75){\line(-1,0){0.2}}\put(5.8,6.75){\line(-1,0){0.2}}\put(5.5,6.75){\line(-1,0){0.2}}\put(5.2,6.75){\line(-1,0){0.2}}\put(4.9,6.75){\line(-1,0){0.2}}\put(8.8,6.75){\line(-1,0){0.2}}\put(9.1,6.75){\line(-1,0){0.2}}\put(9.4,6.75){\line(-1,0){0.2}}\put(9.7,6.75){\line(-1,0){0.2}}\put(10,6.75){\line(-1,0){0.2}}\put(10.3,6.75){\line(-1,0){0.2}}\put(10.35,6.45){ h_2 }\put(4.3,6.45){ h_1 }\put(9.8,7.6){\vector(0,1){0.65}}\put(9.8,7.6){\vector(0,-1){0.85}}\put(9.9,7.45){ h }\put(5.45,7){ R\sin\theta }\end{picture}$

$\blacksquare\:\:\sf{\ \ {h=R\sin\theta+R\cos\theta}}$

$\blacksquare\:\:\sf{\ \ {h_1=R-R\sin\theta}}$

$\blacksquare\:\:\sf{\ \ {h_2=R-R\cos\theta}}$

Now , putting these values in the equation (i)

Now , putting these values in the equation (i)we get ,

$\implies\sf{\ \ {(R-R\sin\theta)\rho_1=(R-R\cos\theta)\rho_1+(R\sin\theta+R\cos\theta)\rho_2}}$

$\implies\sf{\ \ {-\rho_1\sin\theta=-\rho_1\cos\theta+\rho_2\sin\theta+\rho_2\cos\theta}}$

$\implies\sf{\ \ {\rho_1\cos\theta-\rho_1\sin\theta=\rho_2\sin\theta+\rho_2\cos\theta}}$

$\implies\sf{\ \ {\rho_1(\cos\theta-\sin\theta)=\rho_2(\sin\theta+\cos\theta)}}$

$\implies\sf{\ \ {\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\dfrac{\rho_1}{\rho_2}}}$

Divided by $\cos\theta$ , we get ,

$\implies\sf{\ \ {\dfrac{\dfrac{\cos\theta+\sin\theta}{\cos\theta}}{\dfrac{\cos\theta-\sin\theta}{\cos\theta}}=\dfrac{\rho_1}{\rho_2}}}$

$\implies\sf{\ \ {\dfrac{1+\tan\theta}{1-\tan\theta}=\dfrac{\rho_1}{\rho_2}}}$

Adding by 1 on both sides ,

$\implies\sf{\ \ {\dfrac{1+\tan\theta}{1-\tan\theta}+1=\dfrac{\rho_1}{\rho_2}+1}}$

$\implies\sf{\ \ {\dfrac{2}{1-\tan\theta}=\dfrac{\rho_1+\rho_2}{\rho_2}}}$

$\implies\:\:\sf{\ \ {\dfrac{1-\tan\theta}{2}=\dfrac{\rho_2}{\rho_1+\rho_2}}}$

$\implies\sf{\ \ {1-\tan\theta=\dfrac{2\rho_2}{\rho_1+\rho_2}}}$

$\implies\sf{\ \ {\tan\theta=1-\dfrac{2\rho_2}{\rho_1+\rho_2}}}$

$\implies\large{\boxed{\sf{\ \ {\theta=\tan^{-1}\Big(\dfrac{\rho_1-\rho_2}{\rho_1+\rho_2}\Big)}}}}$