A body of mass 2kg is driven by
Question: A body of mass $2 \mathrm{~kg}$ is driven by an engine delivering a constant power of $1 \mathrm{~J} / \mathrm{s}$. The body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in $\mathrm{m}$ )_______ Solution: (18) Given, Mass of the body, $m=2 \mathrm{~kg}$ Power delivered by engine, $P=1 \mathrm{~J} / \mathrm{s}$ Time, $t=9$ seconds Power, $P=F v$ $\Rightarrow P=m a v$ $[\because F=m a]$ $\Rightarrow m \frac{d v}{d t} v=P$ $\left(\because ...
Read More →In an isosceles triangle ABC with AB = AC
Question: In an isosceles triangle ABC with AB = AC and BD AC. Prove that BD2 CD2= 2CD.AD. Solution: Given: ABC is an isosceles triangle with AB = AC and BD AC.To prove: BD2 CD2= 2CD.ADProof: In right ADB, $\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$ (Pythagoras Theorem) $\Rightarrow \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2} \quad(\mathrm{AB}=\mathrm{AC})$ $\Rightarrow(\mathrm{AD}+\mathrm{CD})^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$ $\Rightarrow \mathrm{AD}^{2}+\mathrm{CD}^{2}+2 \mathrm{...
Read More →A body of mass 2kg is driven by
Question: A body of mass $2 \mathrm{~kg}$ is driven by an engine delivering a constant power of $1 \mathrm{~J} / \mathrm{s}$. The body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in $\mathrm{m}$ )_______ Solution: (18) Given, Mass of the body, $m=2 \mathrm{~kg}$ Power delivered by engine, $P=1 \mathrm{~J} / \mathrm{s}$ Time, $t=9$ seconds Power, $P=F v$ $\Rightarrow P=m a v$ $[\because F=m a]$ $\Rightarrow m \frac{d v}{d t} v=P$ $\left(\because ...
Read More →A body of mass 2kg is driven by
Question: A body of mass $2 \mathrm{~kg}$ is driven by an engine delivering a constant power of $1 \mathrm{~J} / \mathrm{s}$. The body starts from rest and moves in a straight line. After 9 seconds, the body has moved a distance (in $\mathrm{m}$ )_______ Solution: (18) Given, Mass of the body, $m=2 \mathrm{~kg}$ Power delivered by engine, $P=1 \mathrm{~J} / \mathrm{s}$ Time, $t=9$ seconds Power, $P=F v$ $\Rightarrow P=m a v$ $[\because F=m a]$ $\Rightarrow m \frac{d v}{d t} v=P$ $\left(\because ...
Read More →Draw two concentric circles of radii 3 cm and 5 cm.
Question: Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on the outer circle, construct the pair of tangents to the inner circle Solution: Steps of ConstructionStep 1. Mark a point O on paper.Step 2. Taking O as a centre, draw two concentric circles of radius 3 cm and 5 cm. Mark any random point P on outer circle.Step 3. Join OP and draw its perpendicular bisector which meets OP at M.Step 4. Draw a circle with M as centre and radius PM (or OM), to intersect the inner circle a...
Read More →Verify the property: x × (y × z) = (x × y) × z by taking:
Question: Verify the property:x (yz) = (xy) zby taking: (i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$ (ii) $x=0, y=\frac{-3}{5}, z=\frac{-9}{4}$ (iii) $x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{5}$ (iv) $x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18}$ Solution: We have to verify that $x \times(y \times z)=(x \times y) \times z$. (i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$ $\mathrm{x} \times(\mathrm{y} \times \mathrm{z})=\frac{-7}{3} \times\left(\frac{12}{5} \times \frac{4}{9}\ri...
Read More →The pair of linear equations 3x + 2y = 5; 2x − 3y = 7 have
Question: The pair of linear equations 3x+ 2y= 5; 2x 3y= 7 have(a) One solution(b) Two solutions(c) Many solutions(d) No solution Solution: The two equations are 3x+ 2y= 5 (1) 2x 3y= 7 (2) Here, $a_{1}=3, b_{1}=2, c_{1}=5$ $a_{2}=2, b_{2}=-3, c_{2}=7$ $\frac{a_{1}}{a_{2}}=\frac{3}{2}, \frac{b_{1}}{b_{2}}=-\frac{2}{3}$ $\therefore \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations....
Read More →Verify the property: x × (y × z) = (x × y) × z by taking:
Question: Verify the property:x (yz) = (xy) zby taking: (i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$ (ii) $x=0, y=\frac{-3}{5}, z=\frac{-9}{4}$ (iii) $x=\frac{1}{2}, y=\frac{5}{-4}, z=\frac{-7}{5}$ (iv) $x=\frac{5}{7}, y=\frac{-12}{13}, z=\frac{-7}{18}$ Solution: We have to verify that $x \times(y \times z)=(x \times y) \times z$. (i) $x=\frac{-7}{3}, y=\frac{12}{5}, z=\frac{4}{9}$ $\mathrm{x} \times(\mathrm{y} \times \mathrm{z})=\frac{-7}{3} \times\left(\frac{12}{5} \times \frac{4}{9}\ri...
Read More →Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.
Question: Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation. Solution: Steps of ConstructionStep 1. Mark a point O on the paper.Step 2. With O as centre and radii 4 cm and 6 cm, draw two concentric circles.Step 3. Mark a point P on the outer circle.Step 4. Join OP.Step 5. Draw the perpendicular bisector XY of OP, cutting OP at Q.Step 6. Draw a circle with Q as centre and r...
Read More →The decimal expansion of
Question: The decimal expansion of $\frac{147}{120}$ will terminate after how many places of decimal? (a) 1(b) 2(c) 3(d) will not terminate Solution: The given number is $\frac{147}{120}$. We have to change the denominator in the form of $2^{m} \times 5^{n}$ and the given number terminates after largest of $m$ and $n$. $\frac{147}{120}=\frac{147}{2^{3} \times 5 \times 3}$ $=\frac{49}{2^{3} \times 5^{1}}$ Here, $m=3, n=1$ Hence the given number terminates after 3 places. Hence the correct option ...
Read More →A bullet of mass 5g,
Question: A bullet of mass $5 \mathrm{~g}$, travelling with a speed of $210 \mathrm{~m} / \mathrm{s}$, strikes a fixed wooden target. One half of its kinetics energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.030 \mathrm{cal} /\left(\mathrm{g}-{ }^{\circ} \mathrm{C}\right)\left(1 \mathrm{cal}=4.2 \times 10^{7}\right.$ ergs $)$ close to :$87.5^{\circ} \mathrm{C}$$83....
Read More →Verify the property: x × y = y × x by taking:
Question: Verify the property:xy=yxby taking: (i) $x=-\frac{1}{3}, y=\frac{2}{7}$ (ii) $x=\frac{-3}{5}, y=\frac{-11}{13}$ (iii) $x=2, y=\frac{7}{-8}$ (iv) $x=0, y=\frac{-15}{8}$ Solution: We have to verify that $x \times y=y \times x$. (i) $x=\frac{-1}{3}, y=\frac{2}{7}$ $x \times y=\frac{-1}{3} \times \frac{2}{7}=\frac{-2}{21}$ $y \times x=\frac{2}{7} \times \frac{-1}{3}=\frac{-2}{21}$ $\therefore \frac{-1}{3} \times \frac{2}{7}=\frac{2}{7} \times \frac{-1}{3}$ Hence verified. (ii) $x=\frac{-3}...
Read More →Draw a circle of radius 3 cm.
Question: Draw a circle of radius 3 cm. Draw a tangent to the circle making an angle of 30 with a line passing through the centre. Solution: Steps Of construction:Step 1. Draw a circle with centre O and radius 3 cm.Step 2. Draw radius OA and produce it to B. Step 3. Make $\angle A O P=60^{\circ}$. Step 4. Draw $\mathrm{PQ} \perp O P$, meeting $\mathrm{OB}$ at $\mathrm{Q}$. Step 5. Then, $\mathrm{PQ}$ is the desired tangent, such that $\angle O Q P=30^{\circ}$....
Read More →A person pushes a box on a rough horizontal platform surface.
Question: A person pushes a box on a rough horizontal platform surface. He applies a force of $200 \mathrm{~N}$ over a distance of $15 \mathrm{~m}$. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to $100 \mathrm{~N}$. The total distance through which the box has been moved is $30 \mathrm{~m}$. What is the work done by the person during the total movement of the hox?$3280 \mathrm{~J}$$2780 \mathrm{~J}$$5690 \mathrm{~J}$$5250 \mathrm{~J}$Correct Option...
Read More →In Fig. 3, AD = 4 cm, BD = 3 cm and CB = 12 cm,
Question: In Fig. 3, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot equals (a) $\frac{3}{4}$ (b) $\frac{5}{12}$ (c) $\frac{4}{3}$ (d) $\frac{12}{5}$ Solution: In Fig 3, if we have AD= 4 cm, BD= 3 cm, and CB= 12 cm, then cot= ? We have the following diagram If we apply Pythagoras theorem in triangleADB, we get hypotenus $=\sqrt{(\text { base })^{2}+(\text { perpendicular })^{2}}$ $\Rightarrow \quad A B=\sqrt{(B D)^{2}+(A D)^{2}}$ $\Rightarrow \quad A B=\sqrt{3^{2}+4^{2}}$ $\Rightarrow \quad A B=5$...
Read More →Write the steps of construction for drawing a pair of tangents to a circle of radius 3 cm,
Question: Write the steps of construction for drawing a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60. Solution: Steps of ConstructionStep 1. Draw a circle with centre O and radius 3 cm.Step 2. Draw any diameter AOB of the circle.Step 3. ConstructBOC = 60 such that radius OC cuts the circle at C.Step 4. Draw AM AB and CN OC. Suppose AM and CN intersect each other at P. Here, AP and CP are the pair of tangents to the circle inclined to each other ...
Read More →Given that tan θ =
Question: Given that $\tan \theta=\frac{1}{\sqrt{3}}$, the value of $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ is (a) 1(b) 1 (c) $\frac{1}{2}$ (d) $-\frac{1}{2}$ Solution: Given: $\tan \theta=\frac{1}{\sqrt{3}}$ We have to find the value of the following expression $\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ Now $\tan \theta=\frac{1}{\sqrt{3}}$, so perpendicular =1 bas...
Read More →Draw a circle of radius 4.2 cm.
Question: Draw a circle of radius 4.2 cm. Draw a pair of tangents to this circle inclined to each other at an angle of 45. Solution: Steps of Construction:Step 1. Draw a circle with centre O and radius = 4.2 cm.Step 2. Draw any diameter AOB of this circle. Step 3. Construct $\angle B O C=45^{\circ}$, such that the radius $O C$ meets the circle at $C$. Step 4. Draw AM $\perp A B$ and $C N \perp O C$. AM and CN intersect at P. Thus, PA and PC are the required tangents to the given circle inclined ...
Read More →Hydrogen ion and singly ionized helium atom are accelerated,
Question: Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to :$1: 2$$10: 7$$2: 1$$5: 7$Correct Option: , 3 Solution: (3) According to work energy theorem, gain in kinetic energy is equal to work done in displacement of charge. $\therefore \quad \frac{1}{2} m v^{2}=q \Delta V$ Here, $\Delta V=$ potential difference between two positions of charge $q$. For same $q$ and $\De...
Read More →Draw a line segment AB of length 8 cm.
Question: Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Solution: Steps of ConstructionStep 1. Draw a line segment AB = 8 cm.Step 2. With A as centre and radius 4 cm, draw a circle.Step 3. With B as centre and radius 3 cm, draw another circle.Step 4. Draw the perpendicular bisector XY of AB, cutting AB at C.Step 5. With C as ...
Read More →Simplify:
Question: Simplify: (i) $\left(\frac{3}{2} \times \frac{1}{6}\right)+\left(\frac{5}{3} \times \frac{7}{2}\right)-\left(\frac{13}{8} \times \frac{4}{3}\right)$ (ii) $\left(\frac{1}{4} \times \frac{2}{7}\right)-\left(\frac{5}{14} \times \frac{-2}{3}\right)+\left(\frac{3}{7} \times \frac{9}{2}\right)$ (iii) $\left(\frac{13}{9} \times \frac{-15}{2}\right)+\left(\frac{7}{3} \times \frac{8}{5}\right)+\left(\frac{3}{5} \times \frac{1}{2}\right)$ (iv) $\left(\frac{3}{11} \times \frac{5}{6}\right)-\left(...
Read More →Draw a circle with the help of a bangle.
Question: Draw a circle with the help of a bangle. Take any point P outside the circle. Construct the pair of tangents from the point P to the circle. Solution: Steps of ConstructionStep 1. Draw a circle with the help of a bangle.Step 2. Mark a point P outside the circle.Step 3. Through P, draw a secant PAB to intersect the circle at A and B.Step 4. Produce AP to C such that PA = PC.Step 5. Draw a semicircle with CB as diameter.Step 6. Draw PD BC, intersecting the semicircle at D.Step 7. With P ...
Read More →A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source.
Question: A particle is moving unidirectionally on a horizontal plane under the action of a constant power supplying energy source. The displacement $(s)$ - time $(t)$ graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale) :Correct Option: , 2 Solution: (2) We know that Power, $P=F v$ But $F=m a v=m \frac{d v}{d t} v$ $\therefore P=m v \frac{d v}{d t} \Rightarrow P d t=m v d v$ Integrating both sides $\int_{0}^{t} P d t=m \int_{0}^{v} v d v$ P. $...
Read More →Draw a circle with centre O and radius 4 cm.
Question: Draw a circle with centre O and radius 4 cm. Draw any diameter AB of this circle. Construct tangents to the circle at each of the two end points of the diameter AB. Solution: Steps of ConstructionStep 1. Draw a circle with centre O and radius 4 cm.Step 2. Draw any diameter AOB of the circle.Step 3. At A, drawOAX = 90. Produce XA to Y.Step 4. At B, drawOBX' = 90. Produce X'B to Y'. Here, XAY and X'BY' are the tangents to the circle at the end points of the diameter AB....
Read More →Draw a circle of radius 3 cm. Take two points P and Q on one of its diameters extended on both sides, each at a distance of 7 cm
Question: Draw a circle of radius 3 cm. Take two pointsPandQon one of its diameters extended on both sides, each at a distance of 7 cm on opposite sides of its centre. Draw tangents to the circle from these two pointsPandQ. Solution: Steps of ConstructionStep 1. Draw a circle with O as centre and radius 3 cm.Step 2. Mark a point P and Q on one of its diameters extended on both sides outside the circle such that OP = OQ = 7 cm.Step 3. Join OP and OQ. Draw the perpendicular bisector XY of OP and X...
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