Mathematics Test - 12

Event A is choosing a yellow pencil first, and Event \(\mathrm{B}\) is choosing a yellow pencil second.

Initially, there are 12 pencils, 7 of which are yellow.

Probability the first pencil is yellow \(=\mathrm{P}(\mathrm{A})=\frac{7}{12}\)

If a yellow pencil is chosen, there will be 11 pencils left, 6 of which are yellow.

Probability the second pencil is yellow \(=\mathrm{P}(\mathrm{B})=\frac{6}{11}\)

Two pencils are chosen at random from the box without replacement.

So events are independent of each other.

Probability they are both yellow:

\(=P(A \cap B)=P(A) \times P(B)\)

\(=\frac{7}{12} \times \frac{6}{11}=\frac{7}{22}\)

Given the length of the string is \(80 \mathrm{~cm}=0.8 \mathrm{~m}\)

It makes \(14\) revolutions in \(25 \mathrm{sec}\)

Frequency is defined as the number of events that are occurring ina given period of time. Its formula is

\(\mathrm{f}= \frac{\text{No. of revolutions }}{\text{time}}\)

Frequency of the string will be \(\mathrm{f}=\frac{14}{25}\)

Now,

Angular speed is defined as the rate of change of angular displacement. In one rotation, angular distance is \(2 \pi\) and the time period is \(T\)

Angular speed is given by formula

\(\Rightarrow \omega=\frac{2 \pi}{t}\)

We know that,

\(\frac{1}{t}=\mathrm{f} \)

\(\Rightarrow \omega=2 \times \pi\times \mathrm{f}\)

Where \(\omega\) is the angular speed \(\mathrm{f}\) is the frequency

Substituting values in the above formula,

\(\Rightarrow \omega=2 \pi \times \frac{14}{25} \)

\(\Rightarrow \omega=\frac{28 \pi}{25} \mathrm{rad} / \mathrm{sec}\)

Also acceleration of a rotating body is given by

\(a=\omega^{2} \mathrm r\)

\(\omega\) is the angular speed of the stone

\(r\) is the radius

\(\Rightarrow a=0.8 \times\left(\frac{28 \pi}{25}\right)^{2} \)

\(\Rightarrow a=0.8 \times \frac{784 \times 9.859}{625}=\frac{6183.564}{625} \)

\(\Rightarrow a=9.89 \mathrm{~m} / \mathrm{s}^{2}\)

So The magnitude of the acceleration of the stone is \(9.89 \mathrm{~m} / \mathrm{s}^{2}\).

Given,

\(\mu=0.\)

\(g=10 \mathrm{~m} / \mathrm{s}^{2}\)

Normal Reaction \((\mathrm{N})=1 \mathrm{~kg} \times 10=10 \mathrm{~N}\)

Friction force will be minimum of applied force or \((\mu \times\) Normal reaction)

Friction force \(\left({F}_{\mathrm{s}}\right)=\) Minimum of \(\{1 \mathrm{~N}, \mu \times \mathrm{N}\}\)

Friction force \(\left(F_{s}\right)=\) Minimum of \(\{1 \mathrm{~N}, 0.3 \times 10\}\)

Friction force \(\left(F_{s}\right)=\) Minimum of \(\{1 \mathrm{~N}, 3 \mathrm{~N}\}\)

Friction force \(\left(F_{s}\right)=1 \mathrm{~N}\)

Given,

The required line passes through the point with position vector \(\hat{i}+2 \hat{j}+\hat{k}\)

\(x_{1}=1, y_{1}=2, z_{1}=1\).

The required line is in the direction of the vector \(\hat{i}-2 \hat{j}+3 \hat{k}\) i.e the required line is parallel to the vector \(\hat{i}-2 \hat{j}+3 \hat{k}\)

\(a=1, b=-2, c=3\).

Now,

We know that,

The cartesian equation of line through a point \(\left(x_{1}, y_{1}, z_{1}\right)\) and having direction ratios \(a, b, c\) is given by:

\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\).

\(\therefore\) The Cartesian equation of the required line is: \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z-1}{3}\)

Given:

\(a+b+c=18, a^{2}+b^{2}+c^{2}=110\) and \(a^{3}+b^{3}+c^{3}=684\)

Formula Used:

\((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)\)

\(a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\)

\(\Rightarrow 18^{2}=110+2(a b+b c+c a)\)

\(\Rightarrow a b+b c+c a=\frac{(324-110)}{2}\)

\(\Rightarrow(a b+b c+c a)=107\) ....(1)

Now,

\(684-3 a b c=18(110-107)\)

\(\Rightarrow 684-3 a b c=54\)

\(\Rightarrow 3 a b c=684-54\)

\(\Rightarrow 3 a b c=630\)

\(\therefore a b c=210\)

Given,

\(x=\frac{\sqrt{3}}{2}\)

\(\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)\)

\(=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}} \times \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)

\(=\frac{(\sqrt{1+x}+\sqrt{1-x})^{2}}{(\sqrt{1+x})^{2}-(\sqrt{1-x})^{2}}\)

\(=\frac{1+x+1-x+2 \sqrt{1-x^{2}}}{1+x-1+x}\)

\(=\frac{2+2 \sqrt{1-x^{2}}}{2 x}\)

\(=\frac{1+\sqrt{1-x^{2}}}{x}\)

\(=\frac{1+\sqrt{1-\frac{3}{4}}}{\sqrt{3}} \times 2\)

\(=\frac{\left(1+\frac{1}{2}\right)}{\sqrt{3}} \times 2\)

\(=\frac{\frac{3}{2} \times 2}{\sqrt{3}}\)

\(=\sqrt{3}\)

Given,

Length of ladder \((A B)=5 \mathrm{~m}, \mathrm{OB}=3 \mathrm{~m}\)

Let \(W\) will be the weight of the ladder, \(N_{B}\) and \(N_{A}\) will be support reaction, \(\theta\) is the angle between ladder and floor and \(\mu\) is the friction coefficient between ladder and floor.

\(O A^{2}=A B^{2}-O B^{2}\)

\( O A^{2}=5^{2}-3^{2}\)

\(OA=4m\)

From \(\triangle \mathrm{OAB}\),

\(\cos \theta=\frac{3}{5}\)

Now apply \(\sum F_{y}=0\)

\(\mathrm{N}_{\mathrm{B}}=\mathrm{W}\)

Now take moment about point \(A\), which should be equal to zero

\(\sum M_{2}=0\)

\(\left(\mu N_{B} \times 4\right)+\left(W \times \frac{5}{2} \times \cos \theta\right)=N_{B} \times 3\)

\(\left(\mu N_{B} \times 4\right)+\left(N_{B} \times \frac{5}{2} \times \frac{3}{5}\right)=N_{B} \times 3\)

\((\mu \times 4)+\left(\frac{3}{2}\right)=3\)

\(\mu=\frac{3}{8}\)

Thus, the value of the coefficient of friction between ladder and floor will be \(\frac{3}{8}\).

Given:

\(f(x, y)=x^{2}-2 x+2 y^{2}+4 y-2\)

Partial derivatives:

\(f^{\prime}(x)=2 x-2\) and \(f^{\prime}(y)=4 y+4\)

Now, for critical points, \(f(x)=0\)

\( 2 x-2=0\)

\( x=1,\)

Also, \(f^{\prime}(y)=0\)

\( 4 y+4=0\)

\( y=-1,\)

So, critical points (1,-1)

\(f^{\prime \prime}(x)=2>0\) and \(f^{\prime}(y)=4>0\)

So, at (1,-1) minimum.