{"id":43774,"date":"2023-05-02T20:08:20","date_gmt":"2023-05-02T14:38:20","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=43774"},"modified":"2023-05-02T20:08:20","modified_gmt":"2023-05-02T14:38:20","slug":"ap-inter-1st-year-physics-important-questions-chapter-4","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-inter-1st-year-physics-important-questions-chapter-4\/","title":{"rendered":"AP Inter 1st Year Physics Important Questions Chapter 4 Motion in a Plane"},"content":{"rendered":"

Students get through AP Inter 1st Year Physics Important Questions<\/a> 4th Lesson Motion in a Plane which are most likely to be asked in the exam.<\/p>\n

AP Inter 1st Year Physics Important Questions 4th Lesson Motion in a Plane<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nThe vertical component of a vector is equal to its horizontal component. What is the angle made by the vector with X-axis? [TS 18; AP 19; lmp.Q]
\nAnswer:
\nVns Let \u2018\u03b8\u2019 be angle made by the vector \\(\\overrightarrow{R}\\) with the X-axis.
\nVertical component of \\(\\overrightarrow{R}\\) = R sin\u03b8
\nHorizontal component of \\(\\overrightarrow{R}\\) = R cos\u03b8
\n\u2234 Rcos\u03b8 = Rsin\u03b8 \u21d2 tan\u03b8 = 1 \u21d2 \u03b8 = 45\u00b0<\/p>\n

Question 2.
\nA vector \\(\\overline{\\mathrm{v}}\\) makes an angle e with the horizontal.
\nThe vector is rotated through an angle e. Does this rotation change the vector \u2018v\\
\nAnswer:
\nBy rotating the vector \\(\\overline{\\mathrm{v}}\\) through angle V it’s magnitude does not change, but its horizontal and vertical components change. Also direction of the vector changes. So the rotation changes the vector \\(\\overline{\\mathrm{v}}\\).<\/p>\n

Question 3.
\nTwo forces of magnitudes 3 units and 5 units act at 60\u00b0 with each other. What is the magnitude of their resultant? |AP 15,16,17||TS 22|
\nAnswer:
\nHere, P = 3 and Q = 5 and \u03b8 = 60\u00b0 \u21d2 cos\u03b8 = cos60\u00b0 = 1\/2
\n\"AP<\/p>\n

Question 4.
\nIf \\(\\overrightarrow{A}\\) = \\(\\overrightarrow{i}+\\overrightarrow{j}\\) What is the angle between vector \\(\\overrightarrow{A}\\) with x -axis? [IPE’ 13, 13, 14; AP 20, 22; TS 17, 20]
\nAnswer:
\nComparing the vector \\(\\overrightarrow{i}+\\overrightarrow{j}\\) with \\(x\\overrightarrow{i}+y\\overrightarrow{j}\\). we get x = 1 and y = 1
\nIf \\(\\overline{\\mathrm{A}}\\) = \\(x\\overline{\\mathrm{i}}+y\\overline{\\mathrm{j}}\\) makes an angle \u03b8 with the x-axis then tan \u03b8 = \\(\\frac{y}{x}=\\frac{1}{1}\\) = 1 (\u2235 \u03b8 = 45\u00b0)<\/p>\n

Question 5.
\nWhen two right-angled vectors of magnitude 7 units and 24 units combine, what is the magnitude of their resultant? [IPE\u2019 14; AP 16,18]
\nAnswer:
\nHerc, P = 7 and Q = 24 and \u03b8 = 90\u00b0 \u21d2 cos\u03b8 = cos90\u00b0 = 0
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 6.
\nIf \\(\\overrightarrow{P}=2\\overrightarrow{i}+4\\overrightarrow{j}+14\\overrightarrow{k}\\) and \\(\\overrightarrow{Q}=4\\overrightarrow{i}+4\\overrightarrow{j}+10\\overrightarrow{k}\\) find the magnitude of P + Q. [TS 15, 16, 19, 22]
\nAnswer:
\n\"AP<\/p>\n

Question 7.
\na vector of magnitude zero have a non zero components?
\nAnswer:
\nNo. A vector of magnitude zero cannot have non-zero components.
\n\"AP<\/p>\n

Question 8.
\nWhat is the acceleration of a projectile at the top of its projectory? [AP, TS 19]
\nAnswer:
\nAt the top of its projectory, the direction of acceleration is vertically downwards and its value is 9.8ms-2<\/sup>.<\/p>\n

Question 9.
\nCan two vectors of unequal magnitude add up to give the zero vector? Can three unequal vectors?
\nAnswer:
\nNo, the addition of two un equal vectors cannot give a zero vector<\/p>\n

But three un equal vectors lying in a plane can give a zero vector, if triangle law is satisfied.<\/p>\n

Short Answer Questions<\/span><\/p>\n

Question 1.
\nState Parallelogram law of vectors. Derive an expression for the magnitude and direction of the resultant vector. [AP 22; TS 16, 17, 20, 22]
\nAnswer:
\nParallelogram law :
\n‘If two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point then their resultant is represented in magnitude and direction by the diagonal passing through the same point\u2019.
\n\"AP
\nLet two vectors \\(\\overrightarrow{P}\\) = \\(\\overrightarrow{OA}\\) and \\(\\overrightarrow{Q}\\) = \\(\\overrightarrow{OB}\\) be acting simultaneously at a point \u2018O\u2019.
\nLet \u2018\u03b8\u2019 be the angle between the two vectors \\(\\overrightarrow{P} and \\overrightarrow{Q}\\) and Q.
\nNow, the parallelogram OACB is completed.
\nThen, the diagonal \\(\\overrightarrow{OC}\\) represents the resultant vector \\(\\overrightarrow{R}\\).
\nNow OA is produced and a perpendicular CD is drawn.
\nThen \u2220CAD = \u03b8
\nAlso, |\\(\\overrightarrow{OA}\\)| = OA = P and |\\(\\overrightarrow{OB}\\)| = OB = AC = Q<\/p>\n

Magnitude of the Resultant vector \\(\\overrightarrow{R}\\) :
\nFrom the right angled triangle COD,
\nOC\u00b2 = OD\u00b2 + CD\u00b2
\n\u21d2 OC\u00b2 = (OA+AD)\u00b2 + CD\u00b2 (Since, OD = OA + AD)
\n\u21d2 OC\u00b2 = OA\u00b2 + AD\u00b2 + 2OA . AD + CD\u00b2 \u21d2 OC\u00b2 = OA\u00b2 + (AD\u00b2 + CD\u00b2) + 2OA . AD ——— (1)
\nNow, from the right angled triangle C AD . AC\u00b2 = AD\u00b2 + CD\u00b2 ——- (2)
\n\"AP<\/p>\n

Direction of the resultant :
\nLet the resultant \\(\\overrightarrow{R}\\) makes an angle ‘\u03b1\u2019 with \\(\\overrightarrow{P}\\).<\/p>\n

\"AP<\/p>\n

Question 2.
\nWhat is relative motion. Explain it?
\nAnswer:
\nRelative motion :
\nThe motion of a body with respect to another body is called relative motion between them.
\n\"AP
\nConsider two bodies A,B moving with Velocities VA<\/sub> and VB<\/sub>.
\n(i) If A and B are moving in the same direction then
\nthe relative velocity of A with respect to B is \\(\\overrightarrow{V}_{AB}=\\overrightarrow{V}_{A}-\\overrightarrow{V}_{B}\\)
\nThe relative velocity of B with respect to A is \\(\\overrightarrow{V}_{BA}=\\overrightarrow{V}_{B}-\\overrightarrow{V}_{A}\\)<\/p>\n

(ii) If A and B are moving in opposite directions then
\nthe relative velocity of A with respect to B is \\(\\overrightarrow{V}_{AB}=\\overrightarrow{V}_{A}-(-\\overrightarrow{V}_{B})=\\overrightarrow{V}_{A}+\\overrightarrow{V}_{B}\\).<\/p>\n

Question 3.
\nShow that a boat must move at an angle of 900 with respect to river water in order to cross the river in a minimum time?
\nAnswer:
\nConsider a boat starts at a point A on one bank of the river and intends to reach the other bank, as shown in the Fig.
\n\"AP<\/p>\n

If the velocity of boat in still water is Vb<\/sub>, and velocity of water in the river is Vw<\/sub> then Resultant velocity is
\n\"AP<\/p>\n

In the above expression, the denominator VR<\/sub> is constant (since Vb<\/sub>, Vw<\/sub> are fixed \/given)
\nHence \u2018t\u2019 becomes minimum when AC becomes minimum. The minimum value of AC is nothing but AB, which is equal to the width \u2018d\u2019 of the river. Here AB is perpendicular to Vw<\/sub>.
\nThus, the boat must move at an angle of 90\u00b0 with respect to the river water.<\/p>\n

Question 4.
\nDefine unit vector, Null vector and position vector [AP 15]
\nAnswer:
\nUnit Vector :
\nA vector of magnitude one unit is called a unit vector.<\/p>\n

Null Vector :
\nA vector of zero magnitude and arbitrary direction is called a zero vector or Null vector.
\n\"AP<\/p>\n

Position Vector :
\nThe position vector of a point P is the vector from the origin \u2018O\u2019 of the coordinate system to the position of the point. It is denoted by \\(\\overrightarrow{OP}\\).<\/p>\n

If (x,y,z) are the co-ordinates of a point ‘P’ and ‘O’ is the origin of coordinate system then \\(\\overrightarrow{OP}=\\overrightarrow{r}=x\\hat{i}+y\\hat{j}+z\\hat{k}\\) is the position vector of P.
\nMagnitude of \\(\\overrightarrow{r}\\) is |\\(\\overrightarrow{r}\\)| = \\(\\sqrt{x^2+y^2+z^2}\\)<\/p>\n

Question 5.
\nIf |\\(\\overrightarrow{a}+\\overrightarrow{b}\\)| = |\\(\\overrightarrow{a}-\\overrightarrow{b}\\)| then what is the angle between \\(\\overrightarrow{a}\\) and \\(\\overrightarrow{b}\\)? [AP 19; TS 18, 22]
\nAnswer:
\n\"AP<\/p>\n

Question 6.
\nShow that the trajectory of an object thrown at a certain angle with the horizontal is a parabola. [AP 20; IPE’ 13,14; AP, TS 15, 16, 17, 18, 19, 22]
\nAnswer:
\n1) Suppose an object is projected from the origin ‘O’ with an initial velocity u at an angle 0 with the horizontal.
\nHorizontal component of u is ux<\/sub> = ucos\u03b8
\nVertical component of u is uy<\/sub> = usin\u03b8<\/p>\n

2) Motion along horizontal: The horizontal component ux<\/sub>= ucos\u03b8 and ax<\/sub> = 0.
\n\"AP<\/p>\n

Thus, (iv) represents the equation of a parabola.
\nSo, the trajectory of a projectile is a parabola.<\/p>\n

\"AP<\/p>\n

Question 7.
\nExplain the terms average velocity and instantaneous velocity. When are they equal?
\nAnswer:
\nAverage velocity :
\nSuppose a body is at a point x1<\/sub> at time t1<\/sub> and at some other point x2<\/sub> at time t2<\/sub>, then the displacement of the body is x2<\/sub> – x1<\/sub>. Hence its average velocity is given by
\n\"AP<\/p>\n

Instantaneous velocity :
\nThe velocity of an object at a particular point of its path or at a particular instant of time is called as instantaneous velocity.
\n\"AP
\nDuring uniform motion, the average and instantaneous velocity are always same because the velocity during uniform motion is same at each point of its path or at each instant.<\/p>\n

Question 8.
\nShow that maximum height and range of projectile are \\(\\frac{u^2 \\sin ^2 \\theta}{2 g}, \\frac{u^2 \\sin 2 \\theta}{g}\\) respectively when the terms have their regular meaning. [Imp.Q|[TS 16]
\nAnswer:
\nMaximum height :
\nIt is the maximum vertical distance travelled by the projectile where its vertical velocity component becomes zero.
\nLet a body be projected with a velocity \u2018u\u2019 at an angle \u2018\u03b8\u2019 with the horizontal.<\/p>\n

In the vertical direction:
\nInitial vertical velocity u = u sin \u03b8
\nAt maximum height, final vertical velocity v = 0
\nAcceleration a = -g.
\nFrom the equation v\u00b2 – u\u00b2= 2as, we get 0 – (usin\u03b8)\u00b2 = -2ghmax<\/sub>
\n\"AP<\/p>\n

Horizontal Range (R) of a projectile :
\nIt is the horizontal distance travelled by a projectile during its time of flight.
\nRange = Horizontal velocity x time of flight
\n\"AP
\nIf \u03b8 is 45\u00b0 then the horizontal range of a projectile is maximum.<\/p>\n

Question 9.
\nIf the trajectory of a body is parabolic in one reference frame, can it be parabolic in another reference frame that moves at constant velocity with respect to the first reference frame? If the trajectory can be other than parabolic, what else it can be?
\nAnswer:
\nIt will not be parabolic from the second reference frame. It appears to move in a straight line.
\nExampler :
\nSuppose a bomb is dropped from an aeroplane flying at a height with a constant speed. For a stationary observer on the earth, the path of the bomb will be parabola. But for the pilot in the plane, the path of the bomb will look like a vertical straight line.<\/p>\n

Question 10.
\nA force \\(2\\hat{i}+\\hat{j}-\\hat{k}\\) newton acts on a body which is initially at rest. At the end of 20 seconds, the velocity of the body is \\(4\\hat{i}+2\\hat{j}-2\\hat{k}\\)ms-1<\/sup>. What is the mass of the body? [AP 16]
\nAnswer:
\n\"AP<\/p>\n

Solved Problems<\/span><\/p>\n

Question 1.
\nRain is falling vertically with a speed of 35ms-1<\/sup>. Wind starts blowing after sometime with a speed of 12 ms-1<\/sup> in east to west direction. In which direction should a boy waiting at a bus stop, hold his umbrella? [Imp.Q]
\nAnswer:
\n\"AP
\nTherefore, the boy should hold his umbrella in the vertical plane at an angle of about 19\u00b0 with the vertical towards the east.<\/p>\n

\"AP<\/p>\n

Question 2.
\nRain is falling vertically with a speed of 35ms-1<\/sup>. A woman rides a bicycle with a speed of 12ms-1<\/sup> in east to west direction. What is the direction in which she should hold her umbrella? [TS 15]
\nAnswer:
\n\"AP
\nThe velocity of rain is vp the velocity of bicycle is vb<\/sub>. Both these velocities are with respect to the ground. Since the woman is riding a bicycle, the velocity of rain as experienced by her is the velocity of rain relative to the velocity of the bicycle. That is
\nvrb<\/sub> = vr<\/sub> – vb<\/sub>
\nThis relative velocity vector makes an angle \u03b8 with the vertical.
\nIt is given by tan \u03b8 = \\(\\frac{v_b}{v_r}=\\frac{12}{35}\\) = 0.343 \u21d2 \u03b8 = Tan-1<\/sup> (0.343) = 19\u00b0<\/p>\n

Therefore, the woman should hold her umbrella at an angle of about 19\u00b0 with the vertical towards the west.<\/p>\n

Question 3.
\nThe position of a particle is given by r = 3.0t\\(\\overline{\\mathrm{i}}\\) + 2.0t\u00b2\\(\\overline{\\mathrm{j}}\\) +5.0\\(\\overline{\\mathrm{k}}\\) where t is in seconds and the coefficients have the proper units for r to be in meters, (a) Find v(t) and a(t) of the particle, (b) Find the magnitude and direction of v(t) at t = 1.0s.
\nSolution:
\n\"AP<\/p>\n

Question 4.
\nProve the statement “for elevations which exceed or fall short of 45\u00b0 by equal amounts, the ranges are equal”. [IMP.Q]
\nSolution:
\nFor a projectile launched with velocity v0 at an angle 90, the range is R = \\(\\frac{\\mathrm{v}_0^2 \\sin \\left(2 \\theta_0\\right)}{\\mathrm{g}}\\)
\nNow, for angles (45\u00b0+ \u03b1) and (45\u00b0- \u03b1) 2\u03b80<\/sub> is (90\u00b0+ 2\u03b1) and (90\u00b0- 2\u03b1) respectively.
\nNow, sin(90\u00b0+2\u03b1) = cos2\u03b1 and sin(90\u00b0-2\u03b1) = cos2\u03b1
\nTherefore, ranges are equal for elevations which exceed or fall short of 45\u00b0 by equal amounts \u03b1.<\/p>\n

Question 5.
\nA cricket ball is thrown at a speed of 28 ms-1<\/sup> in a direction 30\u00b0 above the horizontal. Calculate (a) the maximum height, (b) the time taken by the ball to return to the same level, and (c) the distance from the thrower to the point where the ball returns to the. same level. [IMP.Q]
\nSolution:
\n\"AP<\/p>\n

Exercise Problems<\/span><\/p>\n

Question 1.
\nShip A is 10km due west of ship B. Ship A is heading directly north at a speed of 30 km\/h, while jihip B is heading in a direction 60\u201c west of north at a speed of 20km\/h.
\n(i) Determine the magnitude of the velocity of ship B relative to ship A.
\n(ii) What will be their distance of closest approach?
\nSolution:
\nAssume that ships A and B lie along the X-axis as shown in the diagram.
\n\"AP<\/p>\n

From the concept of relative velocity, ship A may be assumed to be at rest and ship B is moving with a velocity
\n\"AP
\nThe closest distance is the perpendicular distance of the relative velocity vector from the fixed position of ship A.
\n\"AP<\/p>\n

Question 2.
\nIf \u03b8 is angle of projection, R the range, h the maximum height, T the time of flight then . 4h u gT2
\nshow that (a) tan \u03b8 = \\(\\frac{4h}{R}\\) (b) h = \\(\\frac{8T^2}{8}\\).
\nSolution:
\nIf \u03b8 is angle of projection and u is the velocity of projection, then we know that
\n\"AP<\/p>\n

Question 3.
\nA projectile is fired at angle of 60\u00b0 to the horizontal with an initial velocity of 800ms-1.
\n(i) find the time of flight before if hits the ground
\n(ii) find the distance if travels before it hits the ground (range)
\n(iii) find the time of flight for the projectile to reach maximum height.
\nSolution:
\nGiven angle of projection, \u03b8 = 60\u00b0
\nVelocity of projection, u = 800 ms-1<\/sup>.
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 4.
\nFor a particle projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to he \u221a2 limes the maximum height reached by it. Show that the angle of projection is Tan-1<\/sup>(2).
\nSolution:
\nLet P be the maximum height point. Then its coordinates will be (R\/2, h). So its position vector
\n\"AP
\nHence, the angle of projection (\u03b8) = Tan-1<\/sup>(2)<\/p>\n

Question 5.
\nAn object is launched from a cliff 20 m above the ground at an angle of 30\u00b0 above the horizontal with an initial speed of 30 m\/s. How far horizontally does the object travel before landing on the ground? (g=10 m\/s\u00b2)
\nSolution:
\nTotal horizontal distance travelled, R= R1<\/sub> +R2<\/sub>.
\n\"AP
\nAt the point B, in vertical direction, initial velocity (u) = usin\u03b8
\n= 30 \u00d7 sin30\u00b0 = 30 \u00d7 \\(\\frac{1}{2}\\) = 15ms-1<\/sup>
\nAcceleration (a) = g = 10 ms-2<\/sup>; distance travelled (s) = 20m
\nTime taken (t) = ?
\ns = ut + \\(\\frac{1}{2}\\) at\u00b2 \u21d2 20 = 15 \u00d7 t + \\(\\frac{1}{2}\\) \u00d7 10 \u00d7 t\u00b2 \u21d2 20 = 15t + 5t\u00b2 \u21d2 5t\u00b2 + 15t – 20 = 0
\n\u21d2 t\u00b2 + 3t – 4 = 0 \u21d2 t\u00b2 + 4t – t – 4 = 0 \u21d2 t(t+4) – 1(t+4) = 0 \u21d2 (t+4)(t-1) = 0 \u21d2 t =1 (4 cannot be taken)<\/p>\n

Duringthis 1 s, horizontal distance travelled, R2<\/sub> = ucos\u03b8 \u00d7 t = 30 cos30\u00b0 \u00d7 1 = 30 \u00d7 \\(\\frac{\\sqrt{3}}{2}\\) = 15\u221a3m
\n\u2234 Total horizontal distance travelled, R = R1<\/sub> + R2<\/sub> =45\u221a3 + 15\u221a3 =60\u221a3 m<\/p>\n

Question 6.
\nO is a point on the ground chosen as origin. A body first suffers a displacement of 10\u221a2m North-East, next 10m North and finally 10\u221a2m in North-West. How far it is from the origin? [TS 19]
\nSolution:
\n\"AP
\n\u2234 The body is at a distance of 30m from the origin in north direction.<\/p>\n

Question 7.
\nFrom a point on the ground a particle is projected with initial velocity u, such that its horizontal range is maximum. Find the magnitude of average velocity during its ascent.
\nSolustion:
\nFor maximum range, the angle of projection is 45\u00b0.
\nThe body is projected with a velocity u making an angle 45\u00b0 with the horizontal. So u can be
\n\"AP<\/p>\n

Question 8.
\nA particle is projected from the ground with some initial velocity making an angle of 45\u00b0 with the horizontal. It reaches a height of 7:5m above the ground while it travels a horizontal distance of 10m from the point of projection. Find the initial speed of projection. (g = 10m\/s\u00b2).
\nSolution:
\nLet u be the velocity of projection.
\nLet t be the time taken by the body to travel 10m in the horizontal direction.
\nBut we know that, horizontal distance travelled = uniform velocity \u00d7 time
\n\"AP<\/p>\n

Question 9.
\nWind is blowing from the south at 5ms-1<\/sup>. To a cyclist is appears to be blowing from the east at 5ms-1<\/sup>. Show that the velocity of the cyclist is 5\u221a2 ms-1<\/sup> towards north-east.
\nSolution:
\n\"AP
\nVelocity of wind w.r.to ground VWG<\/sub> = 5ms-1<\/sup> from South
\nVelocity of wind w.r.to cyclist, VWG<\/sub> = 5ms-1<\/sup> from East
\nVelocity of cyclist w.r.to ground, VCG<\/sub> = ?
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 10.
\nA person walking at 4 m\/s finds rain drops falling slantwise into his face with a speed of 4m\/s at an angle of 30\u00b0 with the vertical. Show that the actual speed of the rain drops is 4 m\/s.
\nSolution:
\n\"AP
\nVelocity of rain w.r.to man = VRM<\/sub> = 4 ms-1<\/sup>,
\nVelocity of man w.r.to ground = VMG<\/sub> = 4ms-1<\/sup>
\nVelocity of rain w.r.to ground = VRG<\/sub> = ?
\n(actual velocity of rain)
\n\"AP<\/p>\n

Multiple Choice Questions<\/span><\/p>\n

1. If \\(\\overline{\\mathrm{A}}=5\\overline{\\mathrm{i}}+7\\overline{\\mathrm{j}}-3\\overline{\\mathrm{k}}\\) and \\(\\overline{\\mathrm{B}}=2\\overline{\\mathrm{i}}+2\\overline{\\mathrm{j}}-c\\overline{\\mathrm{k}}\\)B-2i + 2j-ck are mutually perpendicular, then the value of ‘c’ is
\n1) 2
\n2) \u221a14
\n3) \u221a10
\n4) \u221a5
\nAnswer:
\n2) \u221a14<\/p>\n

2. \\(\\overline{\\mathrm{A}}=2\\hat{i}-2\\hat{j}+\\hat{k}\\); B = \\(3\\hat{i}+6\\hat{j}+n\\hat{k}\\). What is the value of ‘n’ if \\(\\overline{\\mathrm{A}}\\) and \\(\\overline{\\mathrm{B}}\\) are perpendicular?
\n1) 6
\n2) -6
\n3) 18
\n4) -18
\nAnswer:
\n1) 6<\/p>\n

3. The angle made by the vector \\(\\overline{\\mathrm{A}}=\\overline{\\mathrm{i}}+\\overline{\\mathrm{j}}\\) with X-axis is
\n1) 30\u00b0
\n2) 60\u00b0
\n3) 90\u00b0
\n4) 120\u00b0
\nAnswer:
\n2) 60\u00b0<\/p>\n

4. Force is \\(6\\overline{\\mathrm{i}}+C\\overline{\\mathrm{j}}-2\\overline{\\mathrm{k}}\\) and displacement \\(\\overline{\\mathrm{i}}+2\\overline{\\mathrm{j}}+6\\overline{\\mathrm{k}}\\). If the work done is 6\\(\\overline{\\mathrm{j}}\\), the value of ‘C’ is
\n1) 6
\n2) 4
\n3) 8
\n4) 9
\nAnswer:
\n1) 6<\/p>\n

5. The angle between the vectors (\\(\\overline{\\mathrm{i}}+\\overline{\\mathrm{j}}\\)) and (\\(\\overline{\\mathrm{j}}+\\overline{\\mathrm{k}}\\) + k) is
\n1) 30\u00b0
\n2) 45\u00b0
\n3) 60\u00b0
\n4) 90\u00b0
\nAnswer:
\n3) 60\u00b0<\/p>\n

\"AP<\/p>\n

6. A cricket ball is hit at 45″ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is
\n1) K
\n2) K\/2
\n3) Ksin45\u00b0
\n4) 0
\nAnswer:
\n2) K\/2<\/p>\n

7. The equations of motion of a projectile are given by x = 36t meter and 2y = 96t – 9.8t\u00b2 meter. The angle of projection is
\n\"AP
\nAnswer:
\n1<\/p>\n

8. The speed of a projectile at its maximum height is \u221a3\/2 times its initial speed. If the range of the projectile is p times the maximum height attained by it, then p =
\n1) 4\/3
\n2) 2\u221a3
\n3) 4\u221a3
\n4) 3\/4
\nAnswer:
\n3) 4\u221a3<\/p>\n

9. A person throws a bottle into a dustbin at the same height as he is 2m away at an angle of 45\u00b0. The velocity of thrown bottle is [EAM Q]
\n1) g
\n2) \u221ag
\n3) 2g
\n4) \u221a2g
\nAnswer:
\n4) \u221a2g<\/p>\n

10. Identify the vector quantity among the following.
\n1) Distance
\n2) Angular momentum
\n3) Heat
\n4) Energy
\nAnswer:
\n2) Angular momentum<\/p>\n

11. If a unit vector is represented by \\(0.5\\hat{i}-0.8\\hat{j}+C\\hat{k}\\) then the value of c is
\n\"AP
\nAnswer:
\n2) \u221a0.11<\/p>\n

12. The vectors \\(\\overrightarrow{A}\\) and \\(\\overrightarrow{B}\\) are such that |\\(\\overrightarrow{A}+\\overrightarrow{B}\\)| = |\\(\\overrightarrow{A}-\\overrightarrow{B}\\)|. The angle between the two vectors is
\n1) 45\u00b0
\n2) 90\u00b0
\n3) 60\u00b0
\n4) 75\u00b0
\nAnswer:
\n2) 90\u00b0<\/p>\n

\"AP<\/p>\n

13. If |\\(\\overrightarrow{A}+\\overrightarrow{B}\\)| = \\(|\\overrightarrow{A}|+|\\overrightarrow{B}|\\)| then angle between A and B will be
\n1) 90\u00b0
\n2) 120\u00b0
\n3) 0\u00b0
\n4) 60\u00b0
\nAnswer:
\n3) 0\u00b0<\/p>\n

14. If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is
\n1) 45\u00b0
\n2) 180\u00b0
\n3) 0\u00b0
\n4) 90\u00b0
\nAnswer:
\n4) 90\u00b0<\/p>\n

15. The magnitude of vectors \\(\\overrightarrow{A},\\overrightarrow{B}and\\overrightarrow{C}\\) are 3, 4 and 5 units respectively. If \\(\\overrightarrow{A}+\\overrightarrow{B}=\\overrightarrow{C}\\), the angle between \\(\\overrightarrow{A}and\\overrightarrow{B}\\) is
\n1) \u03c0\/2
\n2) cos-1<\/sup>(0.6)
\n3) tan-1<\/sup>(7\/5)
\n4) \u03c0\/4
\nAnswer:
\n1) \u03c0\/2<\/p>\n

16. A particle has initial velocity (\\(2\\hat{i}+3\\hat{j}\\)) and acceleration (\\(0.3\\hat{i}+0.2\\hat{j}\\)). The magnitude of velocity after 10 seconds will be
\n1) 9\u221a2 units
\n2) 5\u221a2 units
\n3) 5 units
\n4) 9 units
\nAnswer:
\n2) 5\u221a2 units<\/p>\n

17. A particle has initial velocity (\\(3\\hat{i}+4\\hat{j}\\)) and has acceleration (\\(0.4\\hat{i}+0.3\\hat{j}\\)). Its speed after 10 s is
\n1) 7 units
\n2) 7\u221a2 units
\n3) 8.5 units
\n4) 10 units
\nAnswer:
\n2) 7\u221a2 units<\/p>\n

18. The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
\n1) \u03b8 = tan-1<\/sup> (1\/4)
\n2) \u03b8 = tan-1<\/sup>(4)
\n3) \u03b8 = tan-1<\/sup>(2)
\n4) \u03b8 = 45\u00b0
\nAnswer:
\n2) \u03b8 = tan-1<\/sup>(4)<\/p>\n

\"AP<\/p>\n

19. For angles of projection of a projectile at angle (45\u00b0-0) and (45\u00b0+ 0), the horizontal range described by the projectile are in the ratio of
\n1) 2 : 1
\n2) 1 : 1
\n3) 2 : 3
\n4) 1 : 2.
\nAnswer:
\n2) 1 : 1<\/p>\n

20. A missile is fired for maximum range with an initial velocity of 20 m\/s. If g = 10 m\/s\u00b2, the range of the missile is
\n1) 40 m
\n2) 50 m
\n3) 60 m
\n4) 20 m
\nAnswer:
\n1) 40 m<\/p>\n

21. Two projectiles of same mass and with same velocity are thrown at an angle 60\u00b0 and 30\u00b0 with the horizontal, then which will remain same
\n1) time of flight
\n2) range of projectile
\n3) maximum height acquired
\n4) all of them.
\nAnswer:
\n2) range of projectile<\/p>\n

22. The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
\n1) 60\u00b0
\n2) 15\u00b0
\n3) 30\u00b0
\n4) 45\u00b0
\nAnswer:
\n1) 60\u00b0<\/p>\n

23. The maximum range of a gun of horizontal terrain is 16 km. If g = 10 m s-2<\/sup>, then muzzle velocity of a shell must be
\n1) 160 ms-1<\/sup>
\n2) 200\u221a2ms-1<\/sup>
\n3) 400 ms-1<\/sup>
\n4) 800 ms-1<\/sup>
\nAnswer:
\n3) 400 ms-1<\/sup><\/p>\n

24. The velocity of a projectile at the initial point A is (\\(2\\hat{i}+3\\hat{j}\\))m\/s. Its velocity (in m\/s) at point B is
\n1) \\(2\\hat{i}-3\\hat{j}\\)
\n2) \\(2\\hat{i}+3\\hat{j}\\)
\n3) \\(-2\\hat{i}-3\\hat{j}\\)
\n4) \\(-2\\hat{i}+3\\hat{j}\\)
\nAnswer:
\n1) \\(2\\hat{i}-3\\hat{j}\\)<\/p>\n

25. A boat is sent across a river with a velocity of 8 kmh-1<\/sup>. If the resultant velocity of boat is 10 kmh-1<\/sup>, then velocity of river is
\n1) 12.8 kmh-1<\/sup>
\n2) 6 km h-1<\/sup>
\n3) 8 kmh-1<\/sup>
\n4) 10 km h-1<\/sup>
\nAnswer:
\n2) 6 km h-1<\/sup><\/p>\n

\"AP<\/p>\n

26. The width of river is 1 km. The velocity of boat is 5 km\/hr. The boat covered the width of river in shortest time 15 min. Then the velocity of river stream is
\n1) 3 km\/hr
\n2) 4 km\/hr
\n3) \u221a29 km\/hr
\n4) \u221a41 km\/hr
\nAnswer:
\n1) 3 km\/hr<\/p>\n

27. A person aiming to reach exactly opposite point on the hank of a stream is swimming with a speed of 0.5 m\/s at an angle of 120\u00b0 with the direction of flow of water. The speed of water in the stream, is
\n1) 0.25 m\/s
\n2) 0.5 m\/s
\n3) 1.0 m\/s
\n4) 0.433 m\/s
\nAnswer:
\n1) 0.25 m\/s<\/p>\n

28. The speed of a swimmer in still water is 20 m\/s. The speed of river water is 10 m\/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is, given by
\n1) 45\u00b0 west
\n2) 30\u00b0 west
\n3) 0\u00b0
\n4) 60\u00b0 west
\nAnswer:
\n2) 30\u00b0 west<\/p>\n

29. The angular speed of a flywheel making 120 revolutions\/minute is
\n1) 4\u03c0 rad\/s
\n2) 4\u03c0\u00b2 rad\/s
\n3) \u03c0 rad\/s
\n4) 4\u03c0 rad\/s
\nAnswer:
\n1) 4\u03c0 rad\/s<\/p>\n

30. Two particles A and B are moving in uniform circular motion in concentric circles of radii rA<\/sub> and rB<\/sub> with speed vA<\/sub> and vB<\/sub> respectively. Their lime period of rotation is the same. \u2019The ratio of angular speed of A to that of B will be
\n1) 1 : 1
\n2) rA<\/sub> : rB<\/sub>
\n3) vA<\/sub> : vB<\/sub>
\n4) rB<\/sub> : rA<\/sub>
\nAnswer:
\n1) 1 : 1<\/p>\n

31. A particle moves in x-y plane according to rule x = asin\u03c9t and y = acos\u03c9t. The particle follows
\n1) an elliptical path
\n2) a circular path
\n3) a parabolic path
\n4) a straight line path inclined equally to x and y-axes
\nAnswer:
\n2) a circular path<\/p>\n

32. A body is moving with velocity 30 m\/s towards east. After 10 seconds its velocity becomes 40 m\/s towards north. The average acceleration of the body is
\n1) 1 m\/s\u00b2
\n2 ) 7 m\/s\u00b2
\n3) \u221a7 m\/s\u00b2
\n4) 5 m\/s\u00b2
\nAnswer:
\n4) 5 m\/s\u00b2<\/p>\n

33. The x and y coordinates of the particle at any time are x = 5t – 2t\u00b2 and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is
\n1) 5 m s-2<\/sup>
\n2) -4m s-2<\/sup>
\n3) -8 m s-2<\/sup>
\n4) 0
\nAnswer:
\n2) -4m s-2<\/sup><\/p>\n

\"AP<\/p>\n

34. A car starts from rest and accelerates at 5 m\/s\u00b2. At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m\/s\u00b2)
\n1) 20\u221a2 m\/s, 10 m\/s\u00b2
\n2) 20 m\/s, 5 m\/s\u00b2
\n3) 20 m\/s, 0
\n4) 20\u221a2 m \/ s, 0
\nAnswer:
\n1) 20\u221a2 m\/s, 10 m\/s\u00b2<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through AP Inter 1st Year Physics Important Questions 4th Lesson Motion in a Plane which are most likely to be asked in the exam. AP Inter 1st Year Physics Important Questions 4th Lesson Motion in a Plane Very Short Answer Questions Question 1. The vertical component of a vector is equal to its … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[13],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/43774"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=43774"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/43774\/revisions"}],"predecessor-version":[{"id":43853,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/43774\/revisions\/43853"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=43774"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=43774"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=43774"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}