Step 1: Use magnification formula for lenses. \[ m = \frac{-v}{u} \] Step 2: Given that image is formed at twice the distance of object from lens.
So, if object distance = \( u \), then image distance \( v = -2u \) (real and on opposite side).
Step 3: Calculate magnification. \[ m = \frac{-(-2u)}{u} = \frac{2u}{u} = 2 \] So, the magnification is \( 2 \). Since the image is real and inverted, but sign conventions were carefully handled, the magnitude remains positive here.
A current element X is connected across an AC source of emf \(V = V_0\ sin\ 2πνt\). It is found that the voltage leads the current in phase by \(\frac{π}{ 2}\) radian. If element X was replaced by element Y, the voltage lags behind the current in phase by \(\frac{π}{ 2}\) radian.
(I) Identify elements X and Y by drawing phasor diagrams.
(II) Obtain the condition of resonance when both elements X and Y are connected in series to the source and obtain expression for resonant frequency. What is the impedance value in this case?
A country's exports are valued at 800 crore, and its imports are valued at 950 crore in a given year. Due to a trade agreement, the country receives a 10% bonus on its export value from a partner nation. What is the effective trade balance of the country after accounting for the bonus?