100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	If all the letters in the word FIGURES are arranged in alphabetical order from left to right in such a way that vowels are arranged first followed by consonants, then how many letters are there in between U and R after the arrangement?A. twoB. oneC. noneD. three
Answer» Correct Answer - A The given workd is -FIGURES After the rearrangement -EIUFGRS So, the letters between U and R will be two.

Discussion

2.	If in the number 39682147, 1 is added to each of the digit which is less than five and 1 is subtracted from each of the digit which is greather than five then how many digits are repeating in the number thus formed?A. twoB. oneC. noneD. thee
Answer» Correct Answer - B The given number is -39682147 After applied operation -48573256 So, only digit 5 is repeated in the number thus formed.

Discussion

3.	B 5 R 1 @ E K 4 F 7 c L A M 2 P 3 % 9 H I W 8 * 6 U J $ V Q # How many such symbols are there in the above arrangement each of which is immediately preceded by a number but not immediately followed by a consonat?A. NoneB. oneC. TwoD. three
Answer» Correct Answer - D `"1 @,3%,8*"`

Discussion

4.	FIT AZQ TOP RCB SET (The new words formed after performing the mentioned operations may not necessarily be a meaningful English word). If the positions of the first and the third alphabet in each of the words given are interchanged, then how many meaningful word will be formed?A. TwoB. oneC. fourD. three
Answer» Correct Answer - B `{:("FIT AZQ TOP RCB SET"),("TIF QZA POT BCR TES"):}`

Discussion

5.	Seven letters are arranged in a linear arrangement to form a meaningful word. A is second to the left to I. L is to the left of N. Not more than two letters are placed between I and G. G is placed to the right of I. G is not neighbor of E and D. D and E are placed next to each other. Which letter is placed in exactly middle of the meaningful word so formed. If more than one word is formed mark your option as X?A. NB. LC. XD. E
Answer» Correct Answer - C ALIGNED,DEALING

Discussion

6.	FIT AZQ TOP RCB SET (The new words formed after performing the mentioned operations may not necessarily be a meaningful English word). If in each of the word given, the second alphabet is replaced by its following alphabet and third alphabet is replaced by its preceding alphabet as per the English alphabetical order, then how many words thus formed will be without any vowels?A. NoneB. OneC. TwoD. three
Answer» Correct Answer - C `{:("FIT AZQ TOP RCB SET"),("FJS AAP TPO RDA SFS"):}`

Discussion

7.	B 5 R 1 @ E K 4 F 7 c L A M 2 P 3 % 9 H I W 8 * 6 U J $ V Q # Three of the following four are alike in a certain way based on their position in the above arrangement and so form a group. Which is the one that does not belong to that group?A. R1EB. F7LC. M23D. UJ6
Answer» Correct Answer - D UJ6

Discussion

8.	If all the letters of each given words are arranged in alphabetical order within the words, then in how many words are there in which third and fifth letter remains on the same position as earlier? I. MBEKTYD II. GPNAQUS III. XCJRHOA. Only IB. Only I and IIC. All, I,II and IIID. Only II
Answer» Correct Answer - D I MBEKTYD-BDEKMTY II GPNAQSU III. XCJRHO-CHJORX

Discussion

9.	FIT AZQ TOP RCB SET (The new words formed after performing the mentioned operations may not necessarily be a meaningful English word). How many letters are there in the English alphabetical series between the first letter of the word which is second from the left end and the second letter of the word which is third from the right end?A. 11B. 172C. 13D. 15
Answer» Correct Answer - C FIT AZQ TOP RCB SET Hence, there are 13 letters between A and O.

Discussion

10.	FIT AZQ TOP RCB SET (The new words formed after performing the mentioned operations may not necessarily be a meaningful English word). If the given words are arranged in the order as they appear in a dictionary from right to left, which of the following will be second from the left end?A. TOPB. FITC. SETD. AZQ
Answer» Correct Answer - C `{:("FIT AZQ TOP RCB SET"),("TOP SET RCB FIT AZQ"):}`

Discussion

11.	B 5 R 1 @ E K 4 F 7 c L A M 2 P 3 % 9 H I W 8 * 6 U J $ V Q # Which of the following is exactly in the middle between L and U in the above arrangement?A. %B. HC. 9D. 3
Answer» Correct Answer - C 9

Discussion

12.	FIT AZQ TOP RCB SET (The new words formed after performing the mentioned operations may not necessarily be a meaningful English word). If in each of the given words, every consonant is changed to its previous letter and every vowel is changed to its next letter accroding to the English alphabetical series, then in how many words, thus formed, at least one vowels will appear?A. noneB. oneC. twoD. three
Answer» Correct Answer - D `{:("FIT AZQ TOP RCB SET"),("EJS BYP SPO QBA RFS"):}`

Discussion

13.	B 5 R 1 @ E K 4 F 7 c L A M 2 P 3 % 9 H I W 8 * 6 U J $ V Q # Which of the following is the fifth to the left of the seventeenth form the left end of the above arrangement?A. LB. WC. *D. 4
Answer» Correct Answer - A `5^(th)` to left of `17^(th)` from left =17-5 =`12^(th)` from left =L

Discussion

14.	$ F 3 6 N @ 9 K T Q 5 C % 8 B # 7 D S * H 4 W L Step - I The numbers which are immediately preceded by symbol and immediately followed by an alphabet are arranged in the end on the series in increasing order. (Arranged immediate after L) Step -II The odd numbers which are immediately preceded by an alphabet interchange their position with respect to the alphabet just before it. Step -III The alphabets which are immediately followed by a symbol are arranged in alphabetical order between H and 4 of step II. Note: (Step II is applied after Step I and STEP III is applied after STEP II) How many symbols are immediately preceded by alphabets in step III?A. OneB. twoC. threeD. Four
Answer» Correct Answer - B Input $ F 3 6 N @ 9 K T Q 5 C % 8 B # 7 D S * H 4 W L Step I : $ F 3 6 N @ K T Q 5 C % B # D S * H 4 W L 7 8 9 Step II : $ 3 F 6 N @ K T 5 Q C % B # D S * H 4 W 7 L 8 9 Step III : $ 3 F 6 @ K T 5 Q % # D * H B C N S 4 W 7 L 8 9

Discussion

15.	$ F 3 6 N @ 9 K T Q 5 C % 8 B # 7 D S * H 4 W L Step - I The numbers which are immediately preceded by symbol and immediately followed by an alphabet are arranged in the end on the series in increasing order. (Arranged immediate after L) Step -II The odd numbers which are immediately preceded by an alphabet interchange their position with respect to the alphabet just before it. Step -III The alphabets which are immediately followed by a symbol are arranged in alphabetical order between H and 4 of step II. Note: (Step II is applied after Step I and STEP III is applied after STEP II) How many alphabets are immediately preceded and immediately follwed by numbers in step II?A. OneB. twoC. threeD. Four
Answer» Correct Answer - B Input $ F 3 6 N @ 9 K T Q 5 C % 8 B # 7 D S * H 4 W L Step I : $ F 3 6 N @ K T Q 5 C % B # D S * H 4 W L 7 8 9 Step II : $ 3 F 6 N @ K T 5 Q C % B # D S * H 4 W 7 L 8 9 Step III : $ 3 F 6 @ K T 5 Q % # D * H B C N S 4 W 7 L 8 9

Discussion

16.	$ F 3 6 N @ 9 K T Q 5 C % 8 B # 7 D S * H 4 W L Step - I The numbers which are immediately preceded by symbol and immediately followed by an alphabet are arranged in the end on the series in increasing order. (Arranged immediate after L) Step -II The odd numbers which are immediately preceded by an alphabet interchange their position with respect to the alphabet just before it. Step -III The alphabets which are immediately followed by a symbol are arranged in alphabetical order between H and 4 of step II. Note: (Step II is applied after Step I and STEP III is applied after STEP II) What is the sum of number which sixth from right end in step I and eight form left end in step III ?A. 8B. 7C. 9D. 11
Answer» Correct Answer - C Input $ F 3 6 N @ 9 K T Q 5 C % 8 B # 7 D S * H 4 W L Step I : $ F 3 6 N @ K T Q 5 C % B # D S * H 4 W L 7 8 9 Step II : $ 3 F 6 N @ K T 5 Q C % B # D S * H 4 W 7 L 8 9 Step III : $ 3 F 6 @ K T 5 Q % # D * H B C N S 4 W 7 L 8 9

Discussion

17.	A particle is moving on a circular path such that at any instant its position vector, linear velocity. Angular velocity, angular acceleration with respect to centre are `vec(r),vec(v),vec(w),vec(alpha)` respectively. Net acceleration of the particle is `:-`A. `(vec(omega)xxvec(v))-(vec(r)xxvec(alpha))`B. `(vec(omega)xxvec(v))+(vec(r)xxvec(alpha))`C. `(vec(v)xxvec(omega))+(vec(r)xxvec(alpha))`D. `(vec(v)xxvec(omega))-(vec(r)xxvec(alpha))`
Answer» Correct Answer - A

Discussion

18.	If a particle with `a = kv^(2)` and initial velocity is u then velocity after S displacement. Here `k` is a constantA. `ue^(ks)`B. `2ue^(ks)`C. `ue^(2ks)`D. `2ue^(2ks)`
Answer» Correct Answer - A If a particle ………….. `a = (dV)/(dt)` `v(dv)/(ds) = kv^(2)` `int(dv)/(v)=k int ds` `ln v = ks+c` at `s = 0`, `v = u` `implies lnv = ks + lnu` `implies ln (V)/(u)= ks implies v = ue^(ks)`

Discussion

19.	The displacement x in meters of a particle of mass m kg moving in one dimension under the action of a force is related to the time t in seconds by the equation `t=sqrtx+3`, , the work done by the force (in J) in first six seconds is :-A. 18 mB. zeroC. 9 m/2D. 36 m
Answer» Correct Answer - B `t=sqrtx+3` `x=(t-3)^(2)` `x=t^(2)-6t+9` `V=(dx)/(dt)=2t-6` `t=0" "u=-6` `t=6" "v=12-6=6` `W=(1)/(2)xxm\|36-36\|=0`

Discussion

20.	If `vecA=2hati+hatj-hatk. vecB=hati+2hatj+3hatk and vecC=6hati-2hatj-6hatk` then the angle between `(vecA+vecB) and vecC` will be :-A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)`
Answer» Correct Answer - D `vecA+vecB=vecD` `vecD=3hati+3hatj+2hatk" , "vecC=6hati-2hatj-6hatk` `vecDvecC=18-6-12=0` `theta=90^(@)`

Discussion

21.	The least positive value of `x` satisfying the equation `\|x+1\|-\|x\|+3\|x-1\|+2\|x-2\|= x+2` is
Answer» Correct Answer - 1 The least positive value ………… For `0 le x le 1` `x +1-x-3(x-1)-2(x-2)= x+2` `x=1` Hence least positive value of `x = 1`

Discussion

22.	In the figure all spring are identical having spring constant `k` and mass `m` each The block also mass `m` The frequency of oscillation of the block is A. `(1)/(2pi)sqrt((3k)/(m))`B. `(1)/(2pi) sqrt((3k)/(m))`C. `2pi sqrt((3m)/(3k))`D. None of these
Answer» Correct Answer - B In the …………. If spring has the mass then : `f = (1)/(2pi) sqrt((k)/(m+(M_("spring"))/(3)))` Spring equivalent `= 3k` (As all are in parallel) `:. f = (1)/(2pi) sqrt((3k)/(m+((3m))/(3)))` `implies f = (1)/(2pi) sqrt((3k)/(2m))`

Discussion

23.	Illustrated is a uniform cubical block of mass `M` and side `a` Mark the correct statement (s) A. The moment of inertia about axis `A`, passing through the centre of mass is `IA=1/6Ma^(2)`B. The moment of inertia about axis `B`, which bisects one of the cube faces is `IB=5/12Ma^(2)`C. The moment of in nertia about axis `C`, along one of the cube edges is `IC=2/3 Ma^(2)`.D. The moment of inertia about axis `D` which bisects one of the horizontal cube faces is `7/12 Ma^(2)`
Answer» Correct Answer - A::B::C Using the parallel axis theorem, check the distance carefully. `I_D=I_B` (symmetric) `I_A=I_B+m(a/2)^2=5/12Ma^2` `I_C=I_A+M(a/sqrt2)^2=2/3Ma^2`

Discussion

24.	A uniform thin rod is bent in the form of closed loop `ABCDEFA` as shown in the figure. The ratio of moment of inertia of the loop about `x`-axis to that about `y`-axis is A. Greater than `1`B. Less than `1`C. Equal to `1`D. Equal to `0.5`
Answer» Correct Answer - B Moment of inertia of semicircular poritons about X and Y axes same. `I_(X)=I_(Y)=mR^(2)` Then MI of straight portion about X-axis is zero `therefore I_(x)ltI_(y)` (or) `(I_(x))/(I_(Y))lt1`

Discussion

25.	Two particles A and B start simultaneously from the same point and move in a horizontal plane. A has an initial velocity `u_(1)` due east and acceleration `a_(1)` due north. B has an initial velocity `u_(2)` due north and acceleration `a_(2)` due east. Then :A. They must collide at some pointB. They will collide only if `a_(1)u_(1) = a_(2)u_(2)`C. their paths must interesect at same pointD. If `u_(1)gtu_(2) " & " a_(1) lta_(2)`, the particles will have the same speed at some point
Answer» Correct Answer - B::C::D `u_(1)t=(1)/(2)a_(2)t^(2),u_(2)t=(1)/(2)a_(1)t^(2)` `(u_(1))/(u_(2))=(a_(2))/(a_(1))` `u=sqrt(u_(1)^(2)+a_(1)^(2)t^(2))v =sqrt(u_(2)^(2)+a_(2)^(2)t^(2))`

Discussion

26.	An aeroplane is flying vertically upwards. When it is at a height of `1000 m` above the ground and moving at a speed of `367 m//s`., a shot is fired at it with a speed of 567 `ms^(-1)` from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit?A. `gt5 m//s^(2)`B. `gt10 m//s^(2)`C. `lt10 m//s^(2)`D. Not possible
Answer» Correct Answer - B `567t-5t^(2)=1000+367t+(A)/(2)t^(2)` `((A)/(2)+5)t^(2)-200t+1000=0` For not escape `A le 10m//s^(2)` For escape `A gt 10 m//s^(2)`

Discussion

27.	If `vecAxxvecB=vecC`, then choose the incorrect option : [`vecA and vecB` are non zero vectors]A. `vecC` is perpendicular to `(vecA+vecB)`B. `vecC` is perpendicular to `(vecA-vecB)`C. `veC` is perpendicular to `(vecAxxvecB)`D. `vecC` isperpendicular to `vecA` and `vecB`
Answer» Correct Answer - C If `vecA xx vecB = vecC` ……………… `vecC` is perpendicular to `vecA, vecB, (vecA+vecB), (vecA-vecB)` as all these asre in the same plane. `vecC` is parallel to `(vecAxxvecB)`.

Discussion

28.	In `n^(th)` second of the motion, the distance moved by the body is three times the distance moved in the previous second. The motion is uniformly accelerated and the initial velocity was zero. Find n.
Answer» Correct Answer - 2 `S_(n)=u+(1)/(2)a(2n-1)` `S_(n-1)=u+(1)/(2)a(2(n-1)-1)` `S_(n-3)S_(n-1)implies(1)/(2)a(2n-1)=3xx(1)/(2)a(2n-3)` `2n-1=6n-9` `4n=8` n=2

Discussion

29.	If `v = alpha sin (betat)` where v is meter per second and t is in second then. Find sum of dimesions with respect to the time of `alpha` and `beta`A. 1B. `-1`C. 2D. `-2`
Answer» Correct Answer - D `v = alpha sin (beta t)` By dimensional homogenity `alpha` has same dimension of `v = [LT^(-1)] sin (betat)` and `sin (beta t)` no dimension as it is trgonometric function `alpha = M^(0)L^(1)T^(-1)` `beta = M^(0)L^(0)T^(-1)` so w.r.t. = dimensions are `(-1) + (-1) = -2`

Discussion

30.	The acceleration time graph of the motion of a particle from A to B is semicircle with radius of `sqrt(7)` . Find the velocity of particle at `t_(2)` at it `t_(1)` velocity is `3m//sec [pi = 22//7]` . A. `(15)/(2)m//sec`B. `8m//sec`C. `(17)/(2)m//sec`D. None of these
Answer» Correct Answer - C The acceleration ………….. we known that change in velocity = Area under a - t curve. `:. V_(t_(2)) - V_(t_(1))` = Area under the graph and t - axis from D to B (which is quaether circle) `:. V_(t_(2)) - V_(t_(1)) = (pi r^(2))/(4) = (22)/(7)xx((sqrt(7))^(2))/(4) =` `:. V_(t_(2))- 3=(11)/(2) or = (11)/(2) +3 = (17)/(2)m//sec`

Discussion

31.	If \( \frac{d}{d t} \vec{u}=\vec{w} \times \vec{u} \) and \( \frac{d}{d t} \vec{v}=\vec{w} \times \vec{v} \), then show that \( \frac{d}{d t}(\vec{u} \times \vec{v})=\vec{w} \times(\vec{u} \times \vec{v}) \).
Answer» d/dt \(\vec u\) = \(\vec w\) x \(\vec u\) and d/dt \(\vec v\) = \(\vec w\) x \(\vec v\) Let \(\vec u\) = u₁\(\hat i\) + u₂\(\hat j\) + u₃\(\hat k\) and \(\vec v\) = v₁\(\hat i\) + v₂\(\hat j\) + v₃\(\hat k\) \(\vec w\) = w₁\(\hat i\) + w₂\(\hat j\) + w₃\(\hat k\) d\(\vec u\)/dt = du₁/dt \(\hat i\) + du₂/dt \(\hat j\) + du₃/dt \(\hat k\) = \(\vec w\) x \(\vec u\) ⇒ \(\begin{vmatrix}\hat i&\hat j&\hat k\\w_1&w_2&w_3\\u_1&u_2&u_3\end{vmatrix}\) = du₁/dt \(\hat i\) + du₂/dt \(\hat j\) + du₃/dt \(\hat k\) ⇒ \(\hat i\)(w₂ u₃ - u₂ w₃) + \(\hat j\)(w₃ u₁ - u₃ w₁) + \(\hat k\)(w₁u₂ - u₁ w₂) = du₁/dt \(\hat i\) + du₂/dt \(\hat j\) + du₃/dt \(\hat k\) ⇒ du₁/dt = w₂ u₃ - u₂ w₃, du₂/dt = w₃ u₁ - u₃w₁ and du₃/dt = w₁ u₂ - u₁ w₂ Similarly given d\(\vec v\)/dt = \(\vec w\) x \(\vec v\) So we get dv₁/dt = w₂ v₃ - v₂ w₃, dv₂/dt = w₃ v₁ - v₃ w₁ and dv₃/dt = w₁ v₂ - v₁ w₂ L.H.S. = d/dt (\(\vec u\) x \(\vec v\)) = d/dt \(\begin{vmatrix}\hat i&\hat j&\hat k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}\) = d/dt (\(\hat i\)(u₂ v₃ - v₂u₃) + \(\hat j\)(u₃ v₁ - v₃ u₁) + \(\hat k\)(u₁ v₂- v₁ u₂)) = \(\hat i\)(du₂/dt v₃ + u₂ dv₃/dt - dv₂/dt u₃- v₂ du₃/dt) + \(\hat j\)(du₃/dt v₁ + u₃ dv₁/dt - dv₃/dt u₁ - v₃du₁/dt) + \(\hat k\)(du₁/dt v₂ + u₁ dv₂/dt - dv₁/dt u₂ - v₁ du₂/dt) = [\(\hat i\)(du₂/dt u₃ - v₂ du₃/dt) + \(\hat j\)(du₃/dt v₁ - v₃ du₁/dt) + \(\hat k\)(du₁/dt v₂ - v₁ du₂/dt)] + [\(\hat i\)(u₂ dv₃/dt - dv₂/dt u₃) + \(\hat j\)(u₃ dv₁/dt - dv₃/dt u₁) + \(\hat k\)(u₁ dv₂/dt - dv₁/dt u₂)] = (d\(\vec u\)/dt x \(\vec v\)) + (\(\vec u\) x d\(\vec v\)/dt) = (\(\vec w\) x \(\vec u\)) x \(\vec v\)+ (\(\vec u\) x (\(\vec w\) x \(\vec v\))) = -\(\vec v\)x (\(\vec w\) x \(\vec u\)) + \(\vec u\) x (\(\vec w\) x \(\vec v\)) (∵ \(\hat a\) x \(\hat b\) = -\(\hat b\) x \(\hat a\)) = -((\(\vec v\) . \(\vec u\))\(\vec w\) - (\(\vec v\) . \(\vec w\)) . \(\vec u\)) + (\(\vec u\) . \(\vec v\))\(\vec w\) - (\(\vec u\) . \(\vec w\))\(\vec v\) (∵ \(\hat a\) x \(\hat b\) x \(\hat c\) = (\(\hat a\) . \(\hat c\))\(\hat b\) - (\(\hat a\) . \(\hat b\)) \(\hat c\)) = -(\(\vec u\) . \(\vec v\)) \(\vec w\) + (\(\vec v\) . \(\vec w\))\(\vec u\) + (\(\vec u\) . \(\vec v\))\(\vec w\) - (\(\vec u\) . \(\vec w\))\(\vec v\) (∵ \(\hat a\) . \(\hat b\) = \(\hat b\) . \(\hat a\)) = (\(\vec v\) . \(\vec w\))\(\vec u\)) - (\(\vec u\) . \(\vec w\))\(\vec v\) = (\(\vec w\) . \(\vec v\))\(\vec u\) - (\(\vec w\) . \(\vec u\))\(\vec v\) (∵ \(\hat a\) . \(\hat b\) = \(\hat b\) . \(\hat a\)) = \(\vec w\) x (\(\vec u\) x \(\vec v\)) Hence Proved.

31.

If $ \frac{d}{d t} \vec{u}=\vec{w} \times \vec{u} $ and $ \frac{d}{d t} \vec{v}=\vec{w} \times \vec{v} $, then show that $ \frac{d}{d t}(\vec{u} \times \vec{v})=\vec{w} \times(\vec{u} \times \vec{v}) $.

Answer»

d/dt $\vec u$ = $\vec w$ x $\vec u$ and

d/dt $\vec v$ = $\vec w$ x $\vec v$

Let $\vec u$ = u₁$\hat i$ + u₂$\hat j$ + u₃$\hat k$ and

$\vec v$ = v₁$\hat i$ + v₂$\hat j$ + v₃$\hat k$

$\vec w$ = w₁$\hat i$ + w₂$\hat j$ + w₃$\hat k$

d$\vec u$/dt = du₁/dt $\hat i$ + du₂/dt $\hat j$ + du₃/dt $\hat k$ = $\vec w$ x $\vec u$

⇒ $\begin{vmatrix}\hat i&\hat j&\hat k\\w_1&w_2&w_3\\u_1&u_2&u_3\end{vmatrix}$ = du₁/dt $\hat i$ + du₂/dt $\hat j$ + du₃/dt $\hat k$

⇒ $\hat i$(w₂ u₃ - u₂ w₃) + $\hat j$(w₃ u₁ - u₃ w₁) + $\hat k$(w₁u₂ - u₁ w₂)

= du₁/dt $\hat i$ + du₂/dt $\hat j$ + du₃/dt $\hat k$

⇒ du₁/dt = w₂ u₃ - u₂ w₃,

du₂/dt = w₃ u₁ - u₃w₁

and du₃/dt = w₁ u₂ - u₁ w₂

Similarly given d$\vec v$/dt = $\vec w$ x $\vec v$

So we get dv₁/dt = w₂ v₃ - v₂ w₃,

dv₂/dt = w₃ v₁ - v₃ w₁

and dv₃/dt = w₁ v₂ - v₁ w₂

L.H.S. = d/dt ($\vec u$ x $\vec v$) = d/dt $\begin{vmatrix}\hat i&\hat j&\hat k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}$

= d/dt ($\hat i$(u₂ v₃ - v₂u₃) + $\hat j$(u₃ v₁ - v₃ u₁) + $\hat k$(u₁ v₂- v₁ u₂))

= $\hat i$(du₂/dt v₃ + u₂ dv₃/dt - dv₂/dt u₃- v₂ du₃/dt)

+ $\hat j$(du₃/dt v₁ + u₃ dv₁/dt - dv₃/dt u₁ - v₃du₁/dt)

+ $\hat k$(du₁/dt v₂ + u₁ dv₂/dt - dv₁/dt u₂ - v₁ du₂/dt)

= [$\hat i$(du₂/dt u₃ - v₂ du₃/dt) + $\hat j$(du₃/dt v₁ - v₃ du₁/dt) + $\hat k$(du₁/dt v₂ - v₁ du₂/dt)]

+ [$\hat i$(u₂ dv₃/dt - dv₂/dt u₃) + $\hat j$(u₃ dv₁/dt - dv₃/dt u₁) + $\hat k$(u₁ dv₂/dt - dv₁/dt u₂)]

= (d$\vec u$/dt x $\vec v$) + ($\vec u$ x d$\vec v$/dt)

= ($\vec w$ x $\vec u$) x $\vec v$+ ($\vec u$ x ($\vec w$ x $\vec v$))

= -$\vec v$x ($\vec w$ x $\vec u$) + $\vec u$ x ($\vec w$ x $\vec v$) (∵ $\hat a$ x $\hat b$ = -$\hat b$ x $\hat a$)

= -(($\vec v$ . $\vec u$)$\vec w$ - ($\vec v$ . $\vec w$) . $\vec u$) + ($\vec u$ . $\vec v$)$\vec w$ - ($\vec u$ . $\vec w$)$\vec v$

(∵ $\hat a$ x $\hat b$ x $\hat c$ = ($\hat a$ . $\hat c$)$\hat b$ - ($\hat a$ . $\hat b$) $\hat c$)

= -($\vec u$ . $\vec v$) $\vec w$ + ($\vec v$ . $\vec w$)$\vec u$ + ($\vec u$ . $\vec v$)$\vec w$ - ($\vec u$ . $\vec w$)$\vec v$
(∵ $\hat a$ . $\hat b$ = $\hat b$ . $\hat a$)

= ($\vec v$ . $\vec w$)$\vec u$) - ($\vec u$ . $\vec w$)$\vec v$

= ($\vec w$ . $\vec v$)$\vec u$ - ($\vec w$ . $\vec u$)$\vec v$ (∵ $\hat a$ . $\hat b$ = $\hat b$ . $\hat a$)

= $\vec w$ x ($\vec u$ x $\vec v$)

Hence Proved.

Discussion

32.	Show that two non-zero vectors \( \vec{A}=A_{1} \hat{i}+A_{2} \hat{j}+A_{3} \hat{k} \) and \( \vec{B}=B_{1} \hat{i}+B_{2} \hat{j}+B_{3} \hat{k} \) are parallel if and only if \( \frac{A_{1}}{B_{1}}=\frac{A_{2}}{B_{2}}=\frac{A_{3}}{B_{3}} \)
Answer» Vector \(\vec A\) and \(\vec B\) are parallel if and only if \(\vec A\times\vec B=0\) \(\Leftrightarrow\)\(\begin{vmatrix}\hat i&\hat j&\hat k\\A_1&A_2&A_3\\B_1&B_2&B_3\end{vmatrix}=0\) \(\Leftrightarrow\) \(\hat i\) (A₂B₃ - A₃B₂) + \(\hat j\) (B₁A₃ - B₃A₁) + \(\hat k\) (A₁B₂ - B₁A₂) = 0 = 0\(\hat i\) + 0\(\hat j\) + 0\(\hat k\) \(\Leftrightarrow\) A₂B₃ - A₃B₂ = 0, B₁A₃ - B₃A₁ = 0 and A₁B₂ - B₁A₂ = 0 \(\Leftrightarrow\) \(\frac{A_2}{B_2}=\frac{A_3}{B_3},\frac{A_1}{B_1}=\frac{A_3}{B_3}\) and \(\frac{A_1}{B_1}=\frac{A_2}{B_2}\) \(\Leftrightarrow\) \(\frac{A_1}{B_1}=\frac{A_2}{B_2}=\frac{A_3}{B_3}\) Hence Proved.

32.

Show that two non-zero vectors $ \vec{A}=A_{1} \hat{i}+A_{2} \hat{j}+A_{3} \hat{k} $ and $ \vec{B}=B_{1} \hat{i}+B_{2} \hat{j}+B_{3} \hat{k} $ are parallel if and only if $ \frac{A_{1}}{B_{1}}=\frac{A_{2}}{B_{2}}=\frac{A_{3}}{B_{3}} $

Answer»

Vector $\vec A$ and $\vec B$ are parallel if and only if $\vec A\times\vec B=0$

$\Leftrightarrow$$\begin{vmatrix}\hat i&\hat j&\hat k\\A_1&A_2&A_3\\B_1&B_2&B_3\end{vmatrix}=0$

$\Leftrightarrow$ $\hat i$ (A₂B₃ - A₃B₂) + $\hat j$ (B₁A₃ - B₃A₁) + $\hat k$ (A₁B₂ - B₁A₂)

= 0 = 0$\hat i$ + 0$\hat j$ + 0$\hat k$

$\Leftrightarrow$ A₂B₃ - A₃B₂ = 0, B₁A₃ - B₃A₁ = 0 and A₁B₂ - B₁A₂ = 0

$\Leftrightarrow$ $\frac{A_2}{B_2}=\frac{A_3}{B_3},\frac{A_1}{B_1}=\frac{A_3}{B_3}$ and $\frac{A_1}{B_1}=\frac{A_2}{B_2}$

$\Leftrightarrow$ $\frac{A_1}{B_1}=\frac{A_2}{B_2}=\frac{A_3}{B_3}$

Hence Proved.

Discussion

33.	If the point `x_1+t(x_2-x_1), y_1+t(y_2-y_1)` divides the join of `(x_1,y_1)` and `(x_2,y_2)` internally then locus of `t` isA. tlt0B. 0lttlt1C. tgt1D. t=1
Answer» Correct Answer - B `(x_(1)+t(x_(2)-x_(1)),y_(1)+t(y_(2)-y_(1)))` `=(x_(1)(l-t)+tx_(2),y_(1)(l-t)+ty_(2))` is the point which divides the join of `(x_(1) ,y_(1))` and `(x_(2), y_(2))` in the ratio t : (l-t) which is positive if `0lt tlt 1`

Discussion

34.	A common tangent to `9x^2-16y^2 = 144` and `x^2 + y^2 = 9`, isA. `y=(3)/(sqrt(3))X+(15)/(sqrt(7))`B. `y= 3 sqrt((2)/(7))x+(15)/(sqrt(7))`C. `y= 2 sqrt((3)/(7))x+15sqrt(7)`D. None of these
Answer» Correct Answer - B Hyperbola `(x^(2))/(16)-(y^(2))/(9)=1` any tangent to the hy perboile is ` y= mx +-sqrt(a^(2)m^(2)-b^(2))` `implies y= mx +-sqrt(16 m^(2)-9) ....(1)` it toches the circle `implies\|(sqrt(16m^(2)-9))/(sqrt(m^(2)+1))\|=3` `impliesm=3sqrt((2)/(7))` `therefore` Equation of common tangent is `y=3sqrt((2)/(7))x+(15)/(sqrt(7))`

Discussion

35.	Domain of f `(x)=(1)/([X])+sqrt((2-x)x)` is equal to (if [x] denotes the greastest integer function ]A. [0,2]B. [0.1]C. [1,2]D. [1,2]
Answer» Correct Answer - B For the do main `[x]ne 0 and (2-x) ge 0` `x !in (0,1) and x(x-2)le 0`

Discussion

36.	Line 3x + 2y =24 meets x-axis at A and y- axis at B and perpending bisector of AB meets the line passing through (0,1) and parallel to x-axis at C. Area of ` Delta ABC` isA. 182 sq unitsB. 91 sq unitsC. 48 sq unitsD. none of these
Answer» Correct Answer - C `3x+ 2y =24` ` (x)/(8)+(y)/(12)=1` ` A(8,0),b(0,12)` Now perpendicular bisector of AB : 2X - 3Y = `lamda` passes through (4,6) `implies lamda =-10` , ` implies` equation `implies 2x - 3y + 10 =0` `Delta=(1)/(2)\|{:(8,0,1),(0,12,1),((-13)/(2),-1,1):}\|=91`

Discussion

37.	The area of the triangle whose vertices are `A(1,-1,2),B(2,1-1)C(3,-1,2)` is …….A. `sqrt(17)`B. `13`C. `sqrt(13)`D. None of these
Answer» Correct Answer - B Here `bar(OA)= hati-hatj+2hatkand bar(OB)= 2hati+hatj-hatk` `and barc=3hati - hatj+2hatk` `impliesbar(AB)=bar(OB)-bar(OA)=hati +2hatj- 3hatk and ` `bar(AC)=bar(OC)-bar(OA)=2hati` Henve , the required area `=(1)/(2)\|bar(AB)xxbar(AC)\|` `= bar(AB)xxbar(AC)=\|{:(hati,hatj,hatk),(1,2,-3),(2,0,0):}\|=-2(3hatj+2hatk)` `implies` Area of triangle `=(1)/(2)xx2\|3hatj+2hatk\|=sqrt(13)`

Discussion

38.	The three vectors `hati+hatj, hatj+hatk, hatk+hati` taken two at a time form three planes. The three unit vectors drawn perpendicular to these three planes form a parallelopiped of volume.A. 43468B. 4C. `(3 sqrt(3)) 4`D. `4//(3sqrt(3))`
Answer» Correct Answer - C `Let [{:(veca=hati+hatj),(vecb=hatj+hatk),(vecc=hatk+hati):}` `hatr_(1)=((vecaxxvecb))/(\|vecaxxvecb\|)=((hati-hatj+hatk))/(sqrt(3))` `Similarly , hatr_(2)=((vecbxxvecc))/(\|vecbxxvecc\|)=(1)/(sqrt(3))(hati+hatj-hatk)` ` hatr_(3)=((vecbxxvecc))/(\|vecbxxvecc\|)=(1)/(sqrt(3))(-hati+hatj+hatk)` then , the required Volume is `impliesV=(1)/(3sqrt(3))\|{:(1,-1,1),(1,1,-1),(-1,1,1):}\|=(4)/(3sqrt(3))`

Discussion

39.	If tan `4theta = cot 3 theta ` , then general value of `theta`-A. `(2n+1)(pi)/(14),n ne 3k`B. `(2n+1)(pi)/(6)`C. `(2n+1)(pi)/(14)`D. `(2n+1)(pi)/(14),n ne 8 k`
Answer» Correct Answer - C tan `4 theta= cot 3 theta` `thetane(2n+1)(pi)/(8)` `theta ne (npi)/(3)` ` theta =((2n+1)pi)/(2(4+3))` no condition is applicable hence ` theta =((2n+1)pi)/(14)`

Discussion

40.	The equation of the xy - plane is-(a) x = 0,y = 0(b) z = 0(c) x = y ≠ 0(d) none of these
Answer» Answer is (b) z = 0

Discussion

41.	The distance between (4, 3, 7) and (1, -1, -5) is- (a) 13 (b) 15 (c) 12 (d) 5
Answer» Answer is (a) 13

Discussion

42.	If two planes 2x - 4y + 3z = 5 and x + 2y + λz = 12 are perpendicular to each other, then λ = (a) -2 (b) 2 (c) 3 (d) none of these
Answer» Answer is (b) 2

Discussion

43.	Prove that `tan(pi/4+1/2 cos^-1(a/b))+tan(pi/4-1/2 cos^-1(a/b))=(2b)/a`
Answer» Let `1/2cos^-1(a/b) = x` Then, `a/b = cos2x->(1)` Now, `L.H.S. = tan(pi/4+1/2cos^-1(a/b))+tan(pi/4-1/2cos^-1(a/b))` `=tan(pi/4+x)+tan(pi/4-x)` `=(1+tanx)/(1-tanx)+(1-tanx)/(1+tanx)` `=((1+tanx)^2+(1-tanx)^2)/(1-tan^2x)` `=(1+tan^2x+2tanx+1+tan^2x-2tanx)/(1-tan^2x)` `=(2(1+tan^2x))/(1-tan^2x)` We know, `tan^2x = sec^2x-1` So, our expression becomes, `=(2sec^2x)/(2-sec^2x)` `=(2/cos^2x)/(2-1/cos^2x)` `=2/(2cos^2x-1)` `=2/(cos2x)` `=(2)/(a/b)` (From (1)) `=(2b)/a = R.H.S.`

Discussion

44.	Find the coordinates of the point where the line passing through the points (3, 4, 1) and (5, 1, 6) cuts the y - z plane.
Answer» y - z plan is x = 0 ...(i) the eqn. of the line through the point (5, 1, 6) and (3, 4, 1) is (x - 5)/(3 - 5) = (y - 1)/(4 - 1) = (z - 6)/(1 - 6) (x - 5)/-2 = (y - 1)/3 = (z - 6)/-5 ...(ii) Any point an through line is(5 - 2h,1 + 3h,6 - h) Thus f = 8/2 The required point of intersection of plane (i) and two line (ii) is (0,17/2,-13/2)

44.

Find the coordinates of the point where the line passing through the points (3, 4, 1) and (5, 1, 6) cuts the y - z plane.

Answer»

y - z plan is x = 0 ...(i)

the eqn. of the line through the point (5, 1, 6) and (3, 4, 1) is (x - 5)/(3 - 5) = (y - 1)/(4 - 1) = (z - 6)/(1 - 6)

(x - 5)/-2 = (y - 1)/3 = (z - 6)/-5 ...(ii)

Any point an through line is(5 - 2h,1 + 3h,6 - h)

Thus f = 8/2

The required point of intersection of plane (i) and two line (ii) is

(0,17/2,-13/2)

Discussion

45.	Let `A=[(3,2,5),(4,1,3),(0,6,7)]` Express `A` as sum of two matrices such that one is symmetric and the other is skew symmetric.
Answer» `A = [[3,2,5],[4,1,3],[0,6,7]]` Symmetric matrix can be given as `(A+A^T)/2`. Skew-symmetric matrix can be given as `(A-A^T)/2`. Here, `A^T = [[3,4,0],[2,1,6],[5,3,7]]` `:.` Symmetric matrix `=1/2[[6,6,5],[6,2,9],[5,9,14]]` `:.` Skew-symmetric matrix `= 1/2[[0,-2,5],[2,0,-3],[-5,3,0]]` `:. A = 1/2[[6,6,5],[6,2,9],[5,9,14]]+1/2[[0,-2,5],[2,0,-3],[-5,3,0]]`

Discussion

46.	Consider the following statements : Statement I : If A and B be two events corresponding to sample space S such that P(A) = 0.2 and P(B) = 0.8, then A ∪ B is a sure event. Statement II : If A and B are mutually exclusive, then P(A) + P(B) = 1.Of these statements :(a) Both the statements are true and Statement II is the correct explanation of Statement I. (b) Both the statements are true, but statement I is not the correct explanation of Statement I.(c) Statement I is true, but Statement II is false. (d) Statement I is false, but Statement II is true.
Answer» Answer is (a) Both the statements are true and Statement II is the correct explanation of Statement I.

Discussion

47.	Consider the following Statements : Statement I : Let (d/dx){f(x) + c} = F(x) and ∫f(x) dx = f(x) + c. Statement II : Integration is inverse process of differentiation. Of these statements : (a) Both the statements are true and statement II is the correct explanation of Statement I. (b) Both the statements are true, but Statement II is not the correct explanation of Statement I. (c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true.
Answer» Answer is (a) Both the statements are true and statement II is the correct explanation of Statement I.

Discussion

48.	Consider the following statements : Statement I : Let f : N → Y be a function defined as f(x) = 9x + 3, where Y = {y : y = 9x + 3, x ∈ N} then f is one-one.Statement II : For x1,x2 ∈ f(x1) = f(x2) ⇒ x1 = x2Of these statements : (a) Both the statement are true and Statement II is the correct explanation of Statement I. (b) Both the statements are true, but Statement II is not the correct explanation of Statement I.(c) Statement I is true, but Statement II is false.(d) statement I is false, but Statement II is true.
Answer» Answer is (a) Both the statements are true and Statement II is the correct explanation of Statement I.

Discussion

49.	Solve (dy/dx) + 2y tan x = sin x
Answer» The given diff. eqn. (dy/dx) + 2 y tan x = sin x ∴ I.F. = e^{2∫tan x}= e^{2log sec x} = e^{log esec^2x}= sec² x Required soln. y x sec² x = ∫sin x.sec² x dx + c = ∫sec x.tan x dx + c = sec x + c.

49.

Solve (dy/dx) + 2y tan x = sin x

Answer»

The given diff. eqn.

(dy/dx) + 2 y tan x = sin x

∴ I.F. = e^{2∫tan x}= e^{2log sec x} = e^{log esec^2x}= sec² x

Required soln.

y x sec² x = ∫sin x.sec² x dx + c

= ∫sec x.tan x dx + c = sec x + c.

Discussion

50.	Consider the following statements :Statement I : Point of local maximum for f(x) = 4x3 + 9x2 -12x is x = - 2.Statement II : f'(-2) = 0 and f'(-2) > 0.Of these statements : (a) Both the statement, are true and statement II is the correct explanation of Statement I. (b) Both the statements are true, but Statement II is not the correct explanation of Statement I. (c) Statement I is true, but Statement II is false.(d) Statement I is false, but Statement II is true.
Answer» Answer is (b) Both statements are true, but Statement II is not the correct explanation of Statement I.

Discussion

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