The correct answer is: 225
1st refraction:
\(\frac{1.5}{V_1}-0=\frac{0.5}{15}\)
\(⇒ v_1 = 45 \)cm
2nd refraction:
\(\frac{1}{v_2}-\frac{1.5}{15}=\frac{-0.5}{-15}\)
\(⇒ \frac{1}{v_2}=\frac{1}{30}+\frac{1}{10}\)
\(⇒ v2 = +7.5 \)cm
\(⇒\) Distance from centre = 22.5 cm
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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\(F(\frac{dy}{dt},y,t) = 0\)
A partial differential equation is a type, in which the equation carries many unknown variables with their partial derivatives.
It is the linear polynomial equation in which derivatives of different variables exist. Linear Partial Differential Equation derivatives are partial and function is dependent on the variable.
When the degree of f(x,y) and g(x,y) is the same, it is known to be a homogeneous differential equation.
\(\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\)
Read More: Differential Equations