Mathematics Test

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Mathematics Test - 15

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Weekly Quiz Competition

Question 1
4 / -1

A $1 \mathrm{~kg}$ block is resting on a surface with coefficient of friction $0.1$. A force of $0.8 \mathrm{~N}$ is applied to the block as shown in figure. The friction force is:

A
0 N

B
0.8 N

C
0.98 N

D
1.2 N

Solution

Given,

$\mu=0.1$

$ m=1 \mathrm{~kg}$

$\mathrm{F}=0.8 \mathrm{~N}$

Now, we know that

From the FBD as shown below

Normal reaction, $N = mg = 1 × 9.81 = 9.81 N$

Limiting friction force between the block and the surface,

$f=\mu \mathrm{N}$

$=0.1 \times 9.81$

$=0.98 \mathrm{~N}$

But the applied force is $0.8 \mathrm{~N}$ which is less than the limiting friction force.

$\therefore$ The friction force for the given case is $0.8 \mathrm{~N}$.
Question 2
4 / -1

A force F is acting on a bent bar that is clamped at one end as shown in the figure.

A

B

C

D

Solution

When we draw a free body diagram we remove all the supports and force applied due to that support are drawn and force or moment will apply in that manner so that it resists forces in any direction as well as any tendency of rotation.

Following is the conclusion:

Option (B) is wrong because the ground support is shown which we never show in FBD.

Option (C) is wrong because the X component is not shown.

Option (D) is wrong because the moment produced due to the eccentric force is not shown.
Question 3
4 / -1

Two like parallel forces are acting at a distance of $30 \mathrm{~mm}$ apart and their resultant is $60 \mathrm{~N}$, if the line of action of the resultant is $10 \mathrm{~mm}$ from one of the forces, the two forces are:

A
$80 \mathrm{~N}$ and $-20 \mathrm{~N}$

B
$40 \mathrm{~N}$ and $20 \mathrm{~N}$

C
$30 \mathrm{~N}$ and $-90 \mathrm{~N}$

D
$35 \mathrm{~N}$ and $25 \mathrm{~N}$

Solution

Given,

$\mathrm{F}_{1}+\mathrm{F}_{2}=60 \mathrm{~N}$

Taking the moment about the resultant

$F_{1} \times 10-F_{2} \times 20=0$

$F_{1}=2 F_{2}$

By solving, we get

$F_{2}=20 \mathrm{~N}$

$F_{1}=40 \mathrm{~N}$
Question 4
4 / -1

If x, y, z are three consecutive positive integers, then log (1+x z) is:

A
$\log y$

B
$\log \frac{y}{2}$

C
$\log (2 y)$

D
$2 \log (y)$

Solution

Let x, y, z are three consecutive positive integers.

$\therefore y=x+1$ and $z=y+1$ $\Rightarrow z=x+2 $

$\log (1+x z) = \log [1+x(x+2)] $

$=\log \left[1+x^2+2 x\right] $

$ =2 \log y$
Question 5
4 / -1

What is the value of $\lambda$ for which the vectors $2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}$ are perpendicular?

A
$21$

B
$-18$

C
$-22$

D
$22$

Solution

Given: $2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}$ are perpendicular

Let $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}$

We know that, If vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular then $\overrightarrow{a} \cdot \overrightarrow{b}=0$

$\overrightarrow{\mathrm{a}}. \overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-\hat{\mathrm{k}}) \cdot(-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}})=0$

$\Rightarrow-2-20-\lambda=0$

$\Rightarrow-22-\lambda=0$

$\therefore \lambda=-22$
Question 6
4 / -1

A man can swim with a speed of $4.0 \mathrm{~km/h}$ in still water. How long does he take to cross a river $1.0 \mathrm{~km}$ wide if the river flows steadily at $3 \mathrm{~km/h}$ and he makes his strokes normal to the river current? How far does he go in the river when he reaches the other bank of the river?

A
$450 \mathrm{~m}$

B
$690 \mathrm{~m}$

C
$750 \mathrm{~m}$

D
$455 \mathrm{~m}$

Solution

Speed of man $=4.0 \mathrm{~km/h}$

Distance travelled $=1.0 \mathrm{~km}$

Speed of river $=3 \mathrm{~km/h}$

Then, Time $(t)=$ Distance/ Speed

$\frac{1 \mathrm{~km}}{4 \mathrm{~km}}$

$=\frac{1}{4} \mathrm{~h}$

$=\frac{60}{4}$min

$=15$ min

The man is carried down by stream with velocity of river water.

$\therefore$ Distance travelled by man in $15$min (or $\frac{1}{4} \mathrm h$ ) is

$=3 \times \frac{1}{4} \mathrm{~h} $

$=3000 \times \frac{15}{60} $

$=750 \mathrm{~m}$
Question 7
4 / -1

In a GP consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of the GP is:

A
$\frac{1-\sqrt{5}}{2}$

B
$\frac{\sqrt{5}}{2}$

C
$\sqrt{5}$

D
$\frac{\sqrt{5}-1}{2}$

Solution

Let's say that the GP is $a$, ar, ar $^{2}, a r^{3}, \ldots, a r^{n-2}, a r^{n-1},$ with first term a and common ratio r.

Since the GP has only positive terms, it means that $a>0$ and $r>0$.

According to the question, $a r^{n}=a r^{n+1}+a r^{n+2},$ for any $n$. Dividing this equation by $a^{n}$ (which is $>0$ ), we get:

$ 1=r+r^{2}$

$ r^{2}+r-1=0$

$r=\frac{-1 \pm \sqrt{1^{2}-4(1)(-1)}}{2}$

$ \mathrm{r}=\frac{-1+\sqrt{5}}{2}$ OR $\mathrm{r}=\frac{-1-\sqrt{5}}{2}$

Since, $r$ is not negative $(a>0$ and $r>0),$ the common ratio of the series is $\mathrm{r}=$ $\frac{-1+\sqrt{5}}{2}$.
Question 8
4 / -1

If the distance between the points $A(2,0,3)$ and $B(-4, a,-1)$ is $8$ units then find the value of $a$?

A

$\pm 2 \sqrt{5}$

B

$\pm 5 \sqrt{2}$

C

$\pm 3 \sqrt{2}$

D

$\pm 2 \sqrt{3}$

Solution

Given,

$A(2,0,3)$ and $B(-4, a,-1)$ are two points in a $3 D$ space such that distance between them is $8$ units.

As we know that,

If $A\left(x_{1}, y_{1}, z_{1}\right)$ and $B\left(x_{2}, y_{2}, z_{2}\right)$ then the distance between the points $A$ and $B$ is given by:

$\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

$\Rightarrow 8=\sqrt{(-4-2)^{2}+(a-0)^{2}+(-1-3)^{2}}$

By squaring both the sides we get,

$\Rightarrow 64=36+a^{2}+16$

$\Rightarrow a^{2}=12$

$\Rightarrow a=\pm 2 \sqrt{3}$
Question 9
4 / -1

Find the equation of the tangent to the ellipse x² + 2y² = 4 at the point where ordinate is 1 such that point lies in the first quadrant?

A
$\sqrt{2}$ x - 4y - 4 = 0

B
$\sqrt{2}$ x + 4y - 4 = 0

C
$\sqrt{2}$ x + 4y + 4 = 0

D
None of these

Solution

Given:

Equation of ellipse is x² + 2y² = 4

Here, we have to find the equation of the tangent to the given ellipse at the point where ordinate is 1 such that the point lies in the first quadrant.

Let the point be P = (c, 1)

Now this point P lies on the ellipse. So, x = c and y = 1 will satisfy the equation x² + 2y² = 4

c² + 2 = 4

c = ± $\sqrt{2}$

So, the point P is ( $\sqrt{2}$ , 1) because the point P lies in the first quadrant.

The given equation of ellipse we can be re-written as: $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$

As we know that, the equation of the tangent to the horizontal ellipse $\frac{x^{2}}{a^{2}} + \frac{y 2}{b^{2}} = 1$ where 0 < b < a at the point (x₁, y₁) is given by:

$\frac{x \cdot x_{1}}{a^{2}} + \frac{y \cdot y_{1}}{b^{2}} = 1$

$\frac{\sqrt{2} x}{4} + \frac{y}{2} = 1$

$\sqrt{2} x$ + 2y - 4 = 0

So, the required equation of tangent is $\sqrt{2}$ x + 2y - 4 = 0
Question 10
4 / -1

If $y=x^{\frac{1}{ 3}}$ then $\frac{d y}{d x}=?$

A
$\frac{x^{\frac{2}{3}}}{3}$

B
$\frac{1}{3 x^{\frac{2}{3}}}$

C
$x^{-\frac{1}{3}}$

D
$\frac{1}{3 x^{\frac{2}{5}}}$

Solution

Given: $y=x^{\frac{1}{3}}$

Differentiating with respect to $x$, we get:

$\frac{d y}{d x}=\frac{d x^{\frac{1}{3}}}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{3} x^{-\frac{2}{3}} \quad\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{3 x^{\frac{2}{3}}}$