Question

solve it... jaldi

yaad rakhna, spamm is not allowed
vo karega, uska sara answer, questions ka report kar dunga.

★ Any moderator / brainly teacher please help .

who will give correct answer, I give a lot of thanks. ​

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

step 2:

Here, the centre of gravity G lies on the X- axis,

So,

$\bar y \: = 0$

Taking moments about the Y-axis,

$\bar x \sum \: w \: = \sum xw \:\:----(1)$

Conisder the lamina as made up of strips of area a parallel to the Y- axis;

Let the weight par unit area of an element be = p

so, w = p×a = pa

Now from (1) ⇨

$\displaystyle \bar x \sum \: pa\: = \sum xpa \\ \implies \: \bar xp \sum \: a = p \: \sum \: xa \\ \implies \: \bar x \sum \: a \: = \sum xa\: \\ \implies \: \bar x \sum \: 2y \delta x\: = \sum x(2y \delta x) \\ \\ now \: \sum \: = \: \int \: dx \\ \\ \displaystyle \bar x ×\int_0^4\:2y\:dx\:=\:\int_0^4\:2xy \: dx \: \: \: \: \: \{upper \: and \: lower \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: limit \: are \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: collected \: from \: the \: graph \} \\ \implies \: \displaystyle \bar x ×\int_0^4\:2 \times 3 {x}^{ \frac{1}{2} } \:dx\:=\:\int_0^4\:2x \times {x}^{ \frac{1}{2} } \: dx \\ \implies \: \displaystyle \bar x ×\int_0^4\:6 {x}^{ \frac{1}{2} } \:dx\:=\:\int_0^4\:6 {x}^{ \frac{3}{2} } \: dx \\ \implies \: \bar x \times 32 = \frac{384}{5} \\ \implies \: \bar x = \frac{12}{5}$

$So,\: G\:=\:(\:\frac{12}{5}\:,\:0\:)$

Answer 2

12/5 & 0

hope it helps you friend!