- What would be the approximate length of the day if the earth spun so
fast that bodies floated on the equator? Take the radius of the earth R=6\times 10^6 m
and g=9.8 m/s^2
(A) 12 hours (B) 6 hours (C) 3 hours (D) 1.5 hourAnswer: Body remains on earth's surface because
gravitational~ pull ~i.e.~ weight ~of~ the~ body\geq centrifugal~ force ~due ~to~ rotation
Body will start floating when
centrifugal~force~due~ to ~rotation> gravitational~pull~i.e. ~weight~of~the~ body
m\omega^2r>mg or \omega>\sqrt{\frac{g}{r}} But \omega=\frac{2\pi}{T} T(in~ seconds)>2\pi\sqrt{\frac{r}{g}}
T(in~ hours)>2\pi\sqrt{\frac{r}{g}}\frac{1}{3600}=\frac{2\times3.14\times1000\times\sqrt{6}}{3600\times\sqrt{9.8}}=1.5~ hours - The real matrix A=\begin{pmatrix}a&-f&-g\\ f&a&h\\g&-h&a\end{pmatrix} is skew symmetric when
(a) a=0 (b) f=0 (c) g=h (d) f=g Answer: Matrix A is skew symmetric when A_{ij}=-A_{ji}. Hence a=0 - If A and B are matrices such that AB=B and BA=A then A^2+B^2 equals Answer: A^2+B^2=(BA)^2+(AB)^2=BABA+ABAB A^2+B^2=B\underline{AB}A+A\underline{BA}B=B\underline{BA}+A\underline{AB}=BA+AB=A+B
- The average value of function f(x)=4x^3 in the interval 1 to 3 is
Answer: <f(x)>=\frac{\int_1^3f(x)dx}{\int_1^3dx}=\frac{\int_1^34^3dx}{\int_1^3dx}=\frac{[x^4]_1^3}{[x]_1^3}=\frac{81-1}{3-1}=40
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