Question

please solve it fast☹️​

Answer 1

Answer:

$\frac{32}{15}BH^{2}$

Explanation:

$Q= \frac{8B}{\sqrt{H} } \int \limits^H_0 {(H-h)\sqrt{h} } \, dh\\= \frac{8B}{\sqrt{H} } \int \limits^H_0 {(H\sqrt{h} -h \sqrt{h}) } \, dh\\\\= \frac{8B}{\sqrt{H} } \int \limits^H_0 {(H\sqrt{h} -h^{ \frac{3}{2} } ) } \, dh\\\\= \frac{8B}{\sqrt{H} } [\frac{2Hh^{\frac{3}{2} } }{3} -\frac{2 h^{\frac{5}{2} } }{5} ]_{0} ^{H} \\=\frac{8B}{\sqrt{H} } (\frac{2H^{\frac{5}{2} } }{3} - \frac{2H^{\frac{5}{2} } }{5} )\\=\frac{8B}{\sqrt{H} } (2H^{\frac{5}{2} } )(\frac{1}{3} -\frac{1}{5})$

$=\frac{16B}{\sqrt{H} }(H^{\frac{5}{2}} )(\frac{5-3}{15} )\\=\frac{32}{15}BH^{2}$

Formula used:

• $\int\limits^a_b {x^{y} } \, dx \\\\\\= [\frac{x^{y+1} }{y+1} ]_{b} ^{a}\\\\=(\frac{a^{y+1} }{y+1}-\frac{b^{y+1} }{y+1})$

Answer 2

Haan toh yaha dekhlo :)

aur batana kesa hai hehe

niche se padhna sabko