Ask & Answer: India's Largest Education Community

4.32

(a) Let $\mathit{\theta }\left(\mathit{t}\right)=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{{\mathit{v}}_{\mathit{o}\mathit{y}}-\mathit{}\mathit{g}\mathit{t}}{{\mathit{v}}_{\mathit{o}\mathit{x}}}\right)$ be the angle at which the projectile is fired w.r.t. the x-axis, since ${\mathit{\theta }}_0=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{4{\mathit{h}}_{\mathit{m}}}{\mathit{R}}\right)$ depends on t

Therefore tan $\theta$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $\theta$ = $\theta \left(t\right)=\frac{Vx}{Vy}=\frac{Voy-gt}{Vox}$ sin2 $\theta \left(t\right)={\tan }^{-1}?\left(\frac{Voy-gt}{Vox}\right)$ /2g…….(1)

R = $h_{max}$ sin2 $u^2$ /g …….(2)

Dividing (1) by (2)

( $\theta$ /R) = [ $u^2$ sin2 $\theta$ /2g]/[ $h_{max}$ sin2 $u^2$ /g] = $\theta$ / 4

64. Let a and b be the roots of quadratic equation

So, A.M = 8

$\Rightarrow$ $\frac{a+b}{2}=8$ a + b =16 ….I

G.M. = 

$\Rightarrow$ ab = 25…. II

We know that is a quadratic equation

$x^2-x$  (sum of roots) + product of roots = 0

$x^2-x\left(a+b\right)+ab=0$

$x^2-16x+25=0$ using I and II

Which is the reqd. quadratic equation 

Yes, FPM course is available at IMI Bhubaneswar. The FPM course is offered in various specialisations. Candidate must meet the eligibility criteria to enrol for FPM course admission set by the college. Aspirants must pass master degree. IMI Bhubaneswar's admission process is entrance-based.

4.31

Speed of the cyclist = 27 km/h = 7.5 m/s

Radius of the road = 80m

The net acceleration is due to braking and the centripetal acceleration

Acceleration due to braking = 0.5 m/s²

Centripetal acceleration a = v²r = (7.5)2/80= 0.703 m/s²

The resultant acceleration is given by a = sqrt ( $a_t^2$ + $a_c^2$ ) = sqrt ( ${0.5}^2$ + ${0.7}^2$ ) = 0.86 m/s²

tan $\beta$ = AC / at = 0.7/0.5,  $\beta$ = 54.5 $°$

The programme for first three years of the IPM course at IIFT Kakinada is INR 4 lakh per annum. The course fee for fourth and fifth year is as per the prevailing MBA in International Programme fees. The fee excluded additional charges. Candidates have to pay some part of the fee at the time of admission to confirm their seat. Besides, aspirants also need to pay a non-refundable application fee in the start to apply for the course.

4.30

Speed of the fighter plane = 720 km/h = 200 m/s

The altitude of the plane = 1.5 km =1500 m

Velocity of the shell = 600 m/s

$\sin ?\theta$ = 200/600

$\therefore \theta =19.47°$

Let H be the minimum altitude for the plane to fly, without being hit

Using equation H = $u^2$ sin2 (90- $\theta )/2g$

= (6002cos2 $\theta$ )/2g

= 360000 { ( 1 + cos2 $\theta$ )/2}/2 $\times$ 10

= 16000 m = 16 km

For BBA admission, Karnavati University accepts the exam scores of CUET UG and UGAT. It also conducts the Karnavati University Admission Test (KUAT) to shortlist applicants for BBA (Hons) admissions. The exam dates vary from year to year.

4.29 The angle of projectile $\theta$ = 30 $°$

The bullet hits the ground at a distance of 3 km, Range R = 3 km

We know horizontal range for a projectile motion, R = $u^2$ sin2 $\theta$ / g

3 = $u^2$ sin60 $°$ / g

$\frac{u^2}{g}$ = 3/ sin60 $°$ = 3.464 …………… (1)

To hit a target at 5 km,

Max Range,  $R_{max}$ = $\frac{u^2}{g}$ …………. (2)

On comparing equation (1) and (2), we get

$R_{max}$ = 3.464

Hence, the bullet will not hit the target even by adjusting its angle of projection.

Master of Business Administration is one of the best courses offered at the University of the Potomac. With the help of this course, students get to develop communication skills, analytical and strategic thinking abilities, and leadership skills by getting to work in teams. Mentioned below are the details of the University of the Potomac MBA:

Conversion Rate: 1 USD = INR 85.90

Admission to KJSSE BTech course are open, interested candidates can refer to the table below to know the important details related to admission. 

63. Given,

Principal value, amount deposited, P= ?500

Interest Rate, R= 10

Using compound interest = simple interest + $\frac{P\times R\times time}{100}$

Amount at the end of 1st year

= $500+\frac{500\times 10+1}{100}=500\left(1+0.1\right)=500\times 1.1$

= ?500× (1.1)

Amount at the end of 2nd year

= $500\left(1.1\right)+\frac{500\times 1.1\times 10\times 1}{100}$

= $500\left(1.1\right)\left(1+0.1\right)$

= ?500 (1.1)²

Similarly,

Amount at the end of 3rd year = ?500 (1.1)³

So, the amount will form a G.P.

? 500 (1.1)? 500 (1.1)²?500 (1.1)³, ……….

$\therefore$ After 10 years = ?500 (1.1)¹⁰

4.28

(a) We cannot associate the length of a wire bent into a loop with a vector.

(b) We can associate a plane area with a vector.

(c) We cannot associate the volume of a sphere with a vector.

4.27 No in the both the cases.

A physical quantity which is having both direction and magnitude is not necessarily a vector. For instance, in spite of having direction and magnitude, current is a scalar quantity. The basic necessity for a physical quantity to fall in a vector category is that it ought to follow the “law of vector addition”.

As the rotation of a body about an axis does not follow the basic necessity to be a vector i.e. it does not follow the “law of vector addition”.

62. Since the numbers of bacteria doubles every hour. The number after every hour will be a G.P

So, a=30

r=2

$\therefore$ At end of 2^nd hour, a₃ (or 3^rd term) = $a{r}^2=30\times 2^2$

= 30×2⁴

= 120

At end of 4^th hour, a₅ (r 4^th  term) = $a{r}^4$

= 30×2⁴

= 30×16

= 480

Following the trend,

And at the end of nth hour, an+1= $a{r}^{n+1-1}=a{r}^n$

= 30 ×2ⁿ

4.26 Yes and No

A vector in space has no distinct location. The reason behind this is that a vector stays unchanged when it displaces in a way that its direction and magnitude do not change.

A vector changes with time. For instance, the velocity vector of a ball moving with a specific speed fluctuates with time.

Two equivalent vectors situated at different locations in space do not generate the same physical effect. For instance, two equivalent forces acting at different points on a body to rotate the body, but the combination will not generate the equivalent turning effect.

61. Let a and b be the two number and a> b  so a-b = (+ ve)

To get a seat at AURO University, one must satisfy the admission criteria of the desired programme. This includes clearing the eligibility and selection criteria. Students satisfying the eligibility can apply online for the desired course by registering themselves and filling out the application form. Applicants are then shortlisted based on their academic performance, entrance exam scores, PI, or any other selection criteria as specified by the university. Those who make it to the final merit list are then required to go through document verification and pay the admission fees of the university.

Yes, it is possible to get admission at Rajalakshmi Engineering College with 15000 rank. The college accepts TNEA scores for admission to BE/BTech specialisations. The last round closing ranks for the General AI category candidates were between 6193 and 46008. The branch-wise cutoff for the college is given in the table below so that students can know which specialisation is available at which rank:

60. Let a and b be the two numbers and a>b so a-b = (+ ve)

So, sum of two numbers = 6. G.M of a and b

59. We know that,

G.M between a and b = √ab

Given $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\; \surd ab$

$\Rightarrow$ $a^{n+1}+b^{n+1}={\left(ab\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(a^n+b^n\right)$

$\Rightarrow$ $a^{n+1}+b^{n+1}=a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{b}^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]=b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$\Rightarrow$ $\frac{a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}{\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}$

$\Rightarrow$ ${\left(\frac{a}{b}\right)}^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.={\left(\frac{a}{b}\right)}^{?}$

$\Rightarrow$ $n+\frac{1}{2}=0$

$\Rightarrow$ $n=-\frac{1}{2}$