Question

A 20m long brass bar having a cross sectional
area if 400 sq.mm is subjected to axial forces
as shown in figure.
Find the total elongation of the bar. Take
E=2×105 N/mm2.

Answer 1

Explanation:

Given :-

Object distance (u)= -50cm

Radius (R)= -60cm [•°• Concave mirror]

focal length (f)= R/2= -60/2= (-30)cm

Size of object = 15cm

To Find :-

We have to find the size of image

Solution :-

Using mirror formula

\underline{\boxed{\it\ \dfrac{1}{f}= \dfrac{1}{v}+\dfrac{1}{u}}}

f

1

=

v

1

+

u

1

Now , by putting these values in the formula

\begin{gathered}\longrightarrow\sf\ \dfrac{1}{-30}= \dfrac{1}{-50}+\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-1}{30}+\dfrac{1}{50}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-5+3}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ \dfrac{-2}{150}=\dfrac{1}{v}\\ \\ \\ \longrightarrow\sf\ v= \cancel{\dfrac{-150}{2}}\\ \\ \\ \longrightarrow\purple{\sf v= (-75)cm}\end{gathered}

⟶

−30

1

=

−50

1

+

v

1

⟶

30

−1

+

50

1

=

v

1

⟶

150

−5+3

=

v

1

⟶

150

−2

=

v

1

⟶ v=

2

−150

⟶v=(−75)cm

Now we have to find the size of image

By using magnification formula

\underline{\boxed{\sf\ m= \dfrac{-v}{u}= \dfrac{h_i}{h_o}}}

m=

u

−v

=

h

o

h

i

\begin{gathered}\longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{\not{-}(-75)}{\not{-}50}\\ \\ \\ \longrightarrow\sf\ \dfrac{h_i}{15}=\dfrac{-(75)}{50}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-(\cancel{75}\times 15)}{\cancel{50}}\\ \\ \\ \longrightarrow\sf\ \ h_i= \dfrac{-3\times 15}{2}\\ \\ \\ \longrightarrow\sf\ \ h_i= \cancel{\dfrac{-45}{2}}\\ \\ \\ \longrightarrow\sf\ \ h_i= (-22.5)\end{gathered}

⟶

15

h

i

=



−50



−(−75)

⟶

15

h

i

=

50

−(75)

⟶ h

i

=

50

−(

75

×15)

⟶ h

i

=

2

−3×15

⟶ h

i

=

2

−45

⟶ h

i

=(−22.5)

\underline{\bigstar{\textsf{\textbf{\ Size\ of\ image = (-22.5)cm}}}}

★ Size of image = (-22.5)cm