Physics, asked by rupanshu200000, 5 months ago

A block of wood floats in fresh water with2/3 relative density of its volume V submerged and in oil with 0.92 V submerged.Calculate the density of (1)wood,(2) oil.​

Answers

Answered by Anonymous
1

Explanation:

Given: r_{3}=r_{1}+r_{2}+rr

3

=r

1

+r

2

+r

To find: \angle A+\angle B∠A+∠B

Solution:

We know that, for \Delta ABCΔABC ,

r=4R\:sin\frac{A}{2}\:sin\frac{B}{2}\:sin\frac{C}{2}r=4Rsin

2

A

sin

2

B

sin

2

C

r_{1}=4R\:sin\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}r

1

=4Rsin

2

A

cos

2

B

cos

2

C

r_{2}=4R\:cos\frac{A}{2}\:sin\frac{B}{2}\:cos\frac{C}{2}r

2

=4Rcos

2

A

sin

2

B

cos

2

C

r_{3}=4R\:cos\frac{A}{2}\:cos\frac{B}{2}\:sin\frac{C}{2}r

3

=4Rcos

2

A

cos

2

B

sin

2

C

where RR is the circum-radius.

Now, r_{3}=r_{1}+r_{2}+rr

3

=r

1

+r

2

+r

\Rightarrow 4R\:cos\frac{A}{2}\:cos\frac{B}{2}\:sin\frac{C}{2}=4R\:sin\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}+4R\:cos\frac{A}{2}\:sin\frac{B}{2}\:cos\frac{C}{2}+4R\:sin\frac{A}{2}\:sin\frac{B}{2}\:sin\frac{C}{2}⇒4Rcos

2

A

cos

2

B

sin

2

C

=4Rsin

2

A

cos

2

B

cos

2

C

+4Rcos

2

A

sin

2

B

cos

2

C

+4Rsin

2

A

sin

2

B

sin

2

C

\Rightarrow cos\frac{A}{2}\:cos\frac{B}{2}\:sin\frac{C}{2}=sin\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}+cos\frac{A}{2}\:sin\frac{B}{2}\:cos\frac{C}{2}+sin\frac{A}{2}\:sin\frac{B}{2}\:sin\frac{C}{2}⇒cos

2

A

cos

2

B

sin

2

C

=sin

2

A

cos

2

B

cos

2

C

+cos

2

A

sin

2

B

cos

2

C

+sin

2

A

sin

2

B

sin

2

C

\Rightarrow cos\frac{A}{2}\:cos\frac{B}{2}\:sin\frac{C}{2}-cos\frac{A}{2}\:sin\frac{B}{2}\:cos\frac{C}{2}=sin\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}+sin\frac{A}{2}\:sin\frac{B}{2}\:sin\frac{C}{2}⇒cos

2

A

cos

2

B

sin

2

C

−cos

2

A

sin

2

B

cos

2

C

=sin

2

A

cos

2

B

cos

2

C

+sin

2

A

sin

2

B

sin

2

C

\Rightarrow cos\frac{A}{2}\:(cos\frac{B}{2}\:sin\frac{C}{2}-sin\frac{B}{2}\:cos\frac{C}{2})=sin\frac{A}{2}\:(cos\frac{B}{2}\:cos\frac{C}{2}+sin\frac{B}{2}\:sin\frac{C}{2})⇒cos

2

A

(cos

2

B

sin

2

C

−sin

2

B

cos

2

C

)=sin

2

A

(cos

2

B

cos

2

C

+sin

2

B

sin

2

C

)

\Rightarrow -cos\frac{A}{2}\:sin(\frac{B}{2}-\frac{C}{2})=sin\frac{A}{2}\:cos(\frac{B}{2}-\frac{C}{2})⇒−cos

2

A

sin(

2

B

2

C

)=sin

2

A

cos(

2

B

2

C

)

\Rightarrow sin\frac{A}{2}\:cos(\frac{B}{2}-\frac{C}{2})+cos\frac{A}{2}\:sin(\frac{B}{2}-\frac{C}{2})=0⇒sin

2

A

cos(

2

B

2

C

)+cos

2

A

sin(

2

B

2

C

)=0

\Rightarrow sin(\frac{A}{2}+\frac{B}{2}-\frac{C}{2})=0⇒sin(

2

A

+

2

B

2

C

)=0

\Rightarrow sin(\frac{A}{2}+\frac{B}{2}-\frac{C}{2}=sin0⇒sin(

2

A

+

2

B

2

C

=sin0

\Rightarrow \frac{A}{2}+\frac{B}{2}-\frac{C}{2}=0⇒

2

A

+

2

B

2

C

=0

Hope it helpful

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