Image of an object approaching a convex mirror of radius of curvature 20 ~m along its optical axis is observed to move from (25)/(3) ~m to (50)/(7) ~m in 30 ~s . What is the speed of the object in km h^ -1 ?

Correct answer: 3 Using mirror formula twice, (1)/(+25 / 3)+(1)/(-u_1)=(1)/(+10) or (1)/(u_1)=(3)/(25)-(1)/(10) or u_1=50 ~m and, (1)/((+50 / 7))+(1)/(-u_2)=(1)/(+10) (1)/(u_2)=(7)/(50)-(1)/(10) or u_2=25 ~m Speed of object = u_1-u_2 time =(25)/(30) ~ms^ -1 =3 kmh^ -1 The answer is 3 .

JEE Advanced 2010Paper 2

Exam: JEE Advanced
Year: 2010
Paper: Paper 2
Subject: Physics
Chapter: Ray Optics

JEE Advanced 2010 · Paper 2 · Physics · Ray Optics

Image of an object approaching a convex mirror of radius of...

Image of an object approaching a convex mirror of radius of curvature

20 \mathrm{~m}

along its optical axis is observed to move from

\frac{25}{3} \mathrm{~m}

\frac{50}{7} \mathrm{~m}

30 \mathrm{~s}

. What is the speed of the object in

\mathrm{km} \mathrm{h}^{-1}

Correct answer

Step-by-step solution

Using mirror formula twice,

\frac{1}{+25 / 3}+\frac{1}{-u_1}=\frac{1}{+10}

\quad \frac{1}{u_1}=\frac{3}{25}-\frac{1}{10}

u_1=50 \mathrm{~m}

and,

\quad \frac{1}{(+50 / 7)}+\frac{1}{-u_2}=\frac{1}{+10}\therefore \quad \frac{1}{u_2}=\frac{7}{50}-\frac{1}{10}

u_2=25 \mathrm{~m}

Speed of object

=\frac{u_1-u_2}{\text{ time }}=\frac{25}{30} \mathrm{~ms}^{-1}=3 \mathrm{kmh}^{-1}\therefore

The answer is 3 .

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