16 cm
12 cm
To find the focal length of the lens, we use the lens formula:
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. Given that the object distance \(u = -20\) cm (convention: object distance is negative) and the screen is 50 cm away from the object, the image distance \(v = 20 + 50 = 70\) cm.
Substitute these values into the lens formula:
\( \frac{1}{f} = \frac{1}{70} - \frac{1}{-20} \)
\( \frac{1}{f} = \frac{1}{70} + \frac{1}{20} \)
Convert to a common denominator:
\( \frac{1}{f} = \frac{20 + 70}{1400} = \frac{90}{1400} \)
Therefore, \( f = \frac{1400}{90} = \frac{140}{9} \approx 15.56 \text{ cm} \)
So, the focal length of the lens is 16 cm.
The strain-stress plot for materials A, B, C and D is shown in the figure. Which material has the largest Young's modulus? 
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).