Fill in the blanks:
Question: Fill in the blanks: (i) The product of two positive rational numbers is always ..... (ii) The product of a positive rational number and a negative rational number is always ..... (iii) The product of two negative rational numbers is always ..... (iv) The reciprocal of a positive rational number is ..... (v) The reciprocal of a negative rational number is ..... (vi) Zero has ..... reciprocal. (vii) The product of a rational number and its reciprocal is ..... (viii) The numbers ..... and...
Read More →In Fig. 6, AD ⊥ BC and BD =
Question: In Fig. 6, $A D \perp B C$ and $B D=\frac{1}{3} C D$. Prove that $2 C A^{2}=2 A B^{2}+B C^{2}$ Solution: In the given figure, we have $A D \perp B C$ and $B D=\frac{1}{3} C D$, then we have to prove $2 C A^{2}=2 A B^{2}+B C^{2}$ The following given diagram is Now, suppose the value ofBDisx, then $B D=\frac{1}{3} C D$ $\Rightarrow \quad x=\frac{1}{3} C D$ $\Rightarrow C D=3 x$ $B C=B D+C D$ $=x+3 x$ $=4 x$ In triangleADC, we have $C A^{2}=C D^{2}+A D^{2} \ldots \ldots(1)$ And in triangl...
Read More →Construct an isosceles triangle whose base is 9 cm and altitude 5 cm.
Question: Construct an isosceles triangle whose base is $9 \mathrm{~cm}$ and altitude $5 \mathrm{~cm} .$ Construct another triangle whose sides are $\frac{3}{4}$ the corresponding sides of the first isosceles triangle. Solution: Steps of Construction :Step 1. Draw a line segment BC = 9 cm.Step 2. With B as centre, draw an arc each above and below BC.Step 3. With C as centre, draw an arc each above and below BC.Step 4. Join their points of intersection to obtain the perpendicular bisector of BC. ...
Read More →Name the property of multiplication of rational numbers illustrated by the following statements:
Question: Name the property of multiplication of rational numbers illustrated by the following statements: (i) $\frac{-5}{16} \times \frac{8}{15}=\frac{8}{15} \times \frac{-5}{16}$ (ii) $\frac{-17}{5} \times 9=9 \times \frac{-17}{5}$ (iii) $\frac{7}{4} \times\left(\frac{-8}{3}+\frac{-13}{12}\right)=\frac{7}{4} \times \frac{-8}{3}+\frac{7}{4} \times \frac{-13}{12}$ (iv) $\frac{-5}{9} \times\left(\frac{4}{15} \times \frac{-9}{8}\right)=\left(\frac{-5}{9} \times \frac{4}{15}\right) \times \frac{-9}...
Read More →Solve this
Question: Consider a force $\overrightarrow{\mathrm{F}}=-x \hat{i}+y \hat{j}$. The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is: (all quantities are in SI units) 2$\frac{1}{2}$1$\frac{3}{2}$Correct Option: , 3 Solution: (3) Work done, $W=\int \vec{F} \cdot \overrightarrow{d s}$ $=(-x \hat{i}+y \hat{j}) \cdot(d \times \hat{i}+d y \hat{j})$ $\Rightarrow W=-\int_{1}^{0} x d x+\int_{0}^{1} y d y$ $=\left(0+\frac{1}{2}\right)+\frac{1}{2}=1 J$...
Read More →Name the property of multiplication of rational numbers illustrated by the following statements:
Question: Name the property of multiplication of rational numbers illustrated by the following statements: (i) $\frac{-5}{16} \times \frac{8}{15}=\frac{8}{15} \times \frac{-5}{16}$ (ii) $\frac{-17}{5} \times 9=9 \times \frac{-17}{5}$ (iii) $\frac{7}{4} \times\left(\frac{-8}{3}+\frac{-13}{12}\right)=\frac{7}{4} \times \frac{-8}{3}+\frac{7}{4} \times \frac{-13}{12}$ (iv) $\frac{-5}{9} \times\left(\frac{4}{15} \times \frac{-9}{8}\right)=\left(\frac{-5}{9} \times \frac{4}{15}\right) \times \frac{-9}...
Read More →Solve this
Question: Consider a force $\overrightarrow{\mathrm{F}}=-x \hat{i}+y \hat{j}$. The work done by this force in moving a particle from point $A(1,0)$ to $B(0,1)$ along the line segment is: (all quantities are in SI units) 2$\frac{1}{2}$1$\frac{3}{2}$Correct Option: , 3 Solution: (3) Work done, $W=\int \vec{F} \cdot \overrightarrow{d s}$...
Read More →Construct a ∆ABC, in which BC = 5 cm, ∠C = 60° and altitude from A is equal to 3 cm.
Question: Construct a $\triangle A B C$, in which $B C=5 \mathrm{~cm}, \angle C=60^{\circ}$ and altitude from $A$ is equal to $3 \mathrm{~cm}$. Construct a $\triangle A D E$ similar to $\triangle A B C$, such that each side of $\triangle A D E$ is $\frac{3}{2}$ times the corresponding side of ∆ABC.Write the steps of construction. Solution: Steps of Construction :Step 1. Draw a linel.Step 2. Draw an angle of 90oat M onl.Step 3. Cut an arc of radius 3 cm on the perpendicular. Mark the point as A. ...
Read More →Find the multiplicative inverse (reciprocal) of each of the following rational numbers:
Question: Find the multiplicative inverse (reciprocal) of each of the following rational numbers: (i) 9 (ii) 7 (iii) $\frac{12}{5}$ (iv) $\frac{-7}{9}$ (v) $\frac{-3}{-5}$ (vi) $\frac{2}{3} \times \frac{9}{4}$ (vii) $\frac{-5}{8} \times \frac{16}{15}$ (viii) $-2 \times \frac{-3}{5}$ (ix) $-1$ (x) $\frac{0}{3}$ (xi) 1 Solution: (i) Multiplicative inverse (reciprocal) of $9=\frac{1}{9}$ (ii) Multiplicative inverse (reciprocal) of $-7=\frac{-1}{7}$ (iii) Multiplicative inverse (reciprocal) of $\fra...
Read More →In Fig. 5, DE || AC and DF || AE.
Question: In Fig. 5, $\mathrm{DE} \| \mathrm{AC}$ and $\mathrm{DF} \| \mathrm{AE}$. Prove that $\frac{E F}{B F}=\frac{E C}{B E}$ Solution: Given that: IfDE||ACandDF||AE,then we have to prove that $\frac{E F}{B F}=\frac{E C}{B E}$ The following given figure is We can easily see that in the given figure the triangleBDFand triangleBAEare similar triangles and also the triangleBDEand triangleBACare similar triangles. Now we are applying the theorem of similar triangle in triangleBDFand triangleBAE,w...
Read More →Find the multiplicative inverse (reciprocal) of each of the following rational numbers:
Question: Find the multiplicative inverse (reciprocal) of each of the following rational numbers: (i) 9 (ii) 7 (iii) $\frac{12}{5}$ (iv) $\frac{-7}{9}$ (v) $\frac{-3}{-5}$ (vi) $\frac{2}{3} \times \frac{9}{4}$ (vii) $\frac{-5}{8} \times \frac{16}{15}$ (viii) $-2 \times \frac{-3}{5}$ (ix) $-1$ (x) $\frac{0}{3}$ (xi) 1 Solution: (i) Multiplicative inverse (reciprocal) of $9=\frac{1}{9}$ (ii) Multiplicative inverse (reciprocal) of $-7=\frac{-1}{7}$ (iii) Multiplicative inverse (reciprocal) of $\fra...
Read More →A particle (m=1 kg) slides down a frictionless track
Question: A particle $(m=1 \mathrm{~kg})$ slides down a frictionless track (AOC) starting from rest at a point $A$ (height $2 \mathrm{~m}$ ). After reaching $C$, the particle continues to move freely in air as a projectile. When it reaching its highest point $P$ (height $1 \mathrm{~m}$ ), the kinetic energy of the particle (in $J$ ) is: $($ Figure drawn is schematic and not to scale; take $g=10 \mathrm{~ms}^{-2}$ ) Solution: $(10.00)$ Kinetic energy $=$ change in potential energy of the particle...
Read More →Construct a ∆ABC, in which BC = 6.5 cm, AB = 4.5 cm and ∠ABC = 60°.
Question: Construct a $\triangle A B C$, in which $B C=6.5 \mathrm{~cm}, A B=4.5 \mathrm{~cm}$ and $\angle A B C=60^{\circ}$. Construct a triangle similar to this triangle whose sides are $\frac{3}{4}$ the corresponding sides of $\triangle A B C$. Solution: Steps of Construction :Step 1. Draw a line segment BC = 6.5 cm.Step 2. With B as centre, draw an angle of 60o.Step 3. With B as centre and radius equal to 4.5 cm, draw an arc, cutting the angle at A.Step 4. Join AB and AC.Thus, △ ABC is obtai...
Read More →If 7 sin2 θ + 3 cos2θ = 4, show that tan θ
Question: If $7 \sin ^{2} \theta+3 \cos ^{2} \theta=4$, show that $\tan \theta=\frac{1}{\sqrt{3}}$ Solution: Given: $7 \sin ^{2} \theta+3 \cos ^{2} \theta=4$ and we have to prove that $\tan \theta=\frac{1}{\sqrt{3}}$ We can write the given expression as $\left(7 \sin ^{2} \theta\right)+3 \cos ^{2} \theta=4$ $\Rightarrow \quad\left(4 \sin ^{2} \theta+3 \sin ^{2} \theta\right)+3 \cos ^{2} \theta=4$ $\Rightarrow \quad 4 \sin ^{2} \theta+3\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=4$ $\Rightarro...
Read More →Use the distributivity of multiplication of rational numbers over their addition to simplify:
Question: Use the distributivity of multiplication of rational numbers over their addition to simplify: (i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)$ (ii) $\frac{-5}{4} \times\left(\frac{8}{5}+\frac{16}{5}\right)$ (iii) $\frac{2}{7} \times\left(\frac{7}{16}-\frac{21}{4}\right)$ (iv) $\frac{3}{4} \times\left(\frac{8}{9}-40\right)$ Solution: (i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)=\frac{3}{5} \times \frac{35}{24}+\frac{3}{5} \times \frac{10}{1}=\frac{7}{8}+...
Read More →Use the distributivity of multiplication of rational numbers over their addition to simplify:
Question: Use the distributivity of multiplication of rational numbers over their addition to simplify: (i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)$ (ii) $\frac{-5}{4} \times\left(\frac{8}{5}+\frac{16}{5}\right)$ (iii) $\frac{2}{7} \times\left(\frac{7}{16}-\frac{21}{4}\right)$ (iv) $\frac{3}{4} \times\left(\frac{8}{9}-40\right)$ Solution: (i) $\frac{3}{5} \times\left(\frac{35}{24}+\frac{10}{1}\right)=\frac{3}{5} \times \frac{35}{24}+\frac{3}{5} \times \frac{10}{1}=\frac{7}{8}+...
Read More →Draw a line segment AB of length 6.5 cm and divide it in the ratio 4 : 7.
Question: Draw a line segmentABof length 6.5 cm and divide it in the ratio 4 : 7. Measure each of the two parts. Solution: Steps of Construction :Step 1 . Draw a line segment AB = 6.5 cm. Step 2. Draw a ray $A X$, making an acute angle $\angle B A X$. Step 3. Along AX, mark (4+7) =11 points A1, A2,A3,A4,A5, A6, A7, A8, A9, A10, A11,such that AA1= A1A2= A2A3= A3A4= A4A5= A5A6= A6A7= A7A8= A8A9= A9A10= A10A11 Step 4. Join A11B.Step 5. From A4,draw A4C∥A11B, meeting AB at C. Thus, C is the point on...
Read More →In Fig. 4, ABCD is a rectangle. Find the values of x and y.
Question: In Fig. 4,ABCDis a rectangle. Find the values ofxandy. Solution: The following diagram is given The given diagram is rectangle and we know that in rectangle the face to face sides are same. Therefore $x+y=12 \ldots \ldots(1)$ $x-y=8 \ldots \ldots$ (2) Now we are adding the equations (1) and (2) to have $2 x=20$ $\Rightarrow x=10$ Put the value ofxin equation (1), we get $10+y=12$ $\Rightarrow y=12-10$ $\Rightarrow y=2$ Hence,x= 10 andy= 2 Therefore the values of $x$ and $y$ are $x=10, ...
Read More →Verify the property: x × (y + z) = x × y + x × z by taking:
Question: Verify the property:x (y+z) =xy+xzby taking: (i) $x=\frac{-3}{7}, y=\frac{12}{13}, z=\frac{-5}{6}$ (ii) $x=\frac{-12}{5}, y=\frac{-15}{4}, z=\frac{8}{3}$ (iii) $x=\frac{-8}{3}, y=\frac{5}{6}, z=\frac{-13}{12}$ (iv) $x=\frac{-3}{4}, y=\frac{-5}{2}, z=\frac{7}{6}$ Solution: We have to verify that $x \times(y+z)=x \times y+x \times z$. (i) $x=\frac{-3}{7}, y=\frac{12}{13}, z=\frac{-5}{6}$ $\mathbf{x} \times(\mathbf{y}+\mathbf{z})=\frac{-3}{7} \times\left(\frac{12}{13}+\frac{-5}{6}\right)=...
Read More →Can (x − 2) be the remainder on division
Question: Can (x 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer. Solution: If we divide any polynomial by another polynomial, then the degree of divisor is always greater than the degree of remainder. In the given question the degrees of remainder (x2) and divisor (2x+3) are same Therefore if we divide the polynomial P(x) by (2x+3), then (x2) cannot be the remainder...
Read More →If the potential energy between two molecules is given by
Question: If the potential energy between two molecules is given by $U=-\frac{A}{r^{6}}+\frac{B}{r^{12}}$, then at equilibrium, separation between molecules, and the potential energy are:$\left(\frac{B}{2 A}\right)^{\frac{1}{6}},-\frac{A^{2}}{2 B}$$\left(\frac{B}{A}\right)^{\frac{1}{6}}, 0$$\left(\frac{2 B}{A}\right)^{\frac{1}{6}},-\frac{A^{2}}{4 B}$$\left(\frac{2 B}{A}\right)^{\frac{1}{6}},-\frac{A^{2}}{2 B}$Correct Option: , 3 Solution: (3) Given : $U=\frac{-A}{r^{6}}+\frac{B}{r^{12}}$ For equ...
Read More →Determine the value of k so that the following linear equations have no solution:
Question: Determine the value ofkso that the following linear equations have no solution: (3k+ 1)x+ 3y 2 = 0(k2+ 1)x+ (k 2)y 5 = 0 Solution: The given system of equations is(3k+ 1)x+ 3y 2 = 0(k2+ 1)x+ (k 2)y 5 = 0Here,a1= 3k+ 1,b1= 3,c1= 2a2=k2+ 1,b2=k 2,c2= 5The given system of equations has no solution. $\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ $\Rightarrow \frac{3 k+1}{k^{2}+1}=\frac{3}{k-2} \neq \frac{-2}{-5}$ $\Rightarrow \frac{3 k+1}{k^{2}+1}=\frac{3}{k-...
Read More →Draw a line segment AB of length 5.4 cm.
Question: Draw a line segment AB of length 5.4 cm. Divide it into six equal parts. Write the steps of construction. Solution: Steps of Construction :Step 1 . Draw a line segment AB = 5.4 cm. Step 2. Draw a ray $A X$, making an acute angle, $\angle B A X$. $\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}=\mathrm{A}_{3} \mathrm{~A}_{4}=\mathrm{A}_{4} \mathrm{~A}_{5}=\mathrm{A}_{5} \mathrm{~A}_{6}$. Step 4. Join A6B.Step 5. Draw A1C∥A2D, A3D, A4F and A5G . Thus, AB is ...
Read More →Write the steps of construction to construct the tangents to a circle from an external point.
Question: Write the steps of construction to construct the tangents to a circle from an external point. Solution: Steps of ConstructionStep 1. Draw a circle with O as centre and some radius.Step 2. Mark a point P outside the circle.Join OP.Step 3. Draw the perpendicular bisector of OP cutting it at M.Step 4. Draw another circle with M as centre and radius MP (or OM), to intersect the given circle at the points A and B.Step 5. Join PA and PB. Here, PA and PB are the required tangents....
Read More →For a given data with 70 observations the 'less then ogive'
Question: For a given data with 70 observations the 'less then ogive' and the 'more than ogive' intersect at (20.5, 35). The median of the data is(a) 20(b) 35(c) 70(d) 20.5 Solution: It is given that less than ogive and more than ogive intersects at (20.5, 35), then we have to find the median. We know that for a given distribution, median is thexcoordinates of intersection of less than ogive curve and more than ogive curve. Since the intersection point is (20.5, 35), therefore the median is 20.5...
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