141357
Two cars \(P\) and \(Q\) start from a point at the same time in a straight line and their positions are represented by \(X_{P}(t)=a t+b t^{2}\) and \(X_{Q}(t)=\) \(\mathrm{ft}-\mathrm{t}^{2}\). At what time do the cars have the same velocity?
D Given: \(X_{P}(t)=a t+b t^{2} \ldots\) (i) \(\text { And } X_{Q}(t)=f t-t^{2}\) Differentiating eqn (i) and (ii) w.r.t ' \(t\) ' we get, \(\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{P}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{at}+\mathrm{bt}^{2}\right)}{\mathrm{dt}}=\mathrm{a}+2 \mathrm{bt}\) \(\mathrm{v}_{\mathrm{Q}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{Q}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{ft}-\mathrm{t}^{2}\right)}{\mathrm{dt}}=\mathrm{f}-2 \mathrm{t}\) Let at time ' \(\mathrm{t}_{0}\) ' the cars have same velocity \({\left[\mathrm{v}_{\mathrm{P}}\right]_{\mathrm{t}=\mathrm{t}_{0}}=\left[\mathrm{v}_{\mathrm{Q}}\right]_{\mathrm{t}=\mathrm{t}_{0}}}\) \(\Rightarrow \mathrm{a}+2 \mathrm{bt}_{0}=\mathrm{f}-2 \mathrm{t}_{0}\) \(\Rightarrow \mathrm{t}_{0}=\frac{\mathrm{f}-\mathrm{a}}{2(1+\mathrm{b})}\)
NEET 2016
Motion in One Dimensions
141358
The displacement-time graphs of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(x\) axis as shown in the figure. The ratio of their respective velocity is
1 \(1: 2\)
2 \(1: \sqrt{3}\)
3 \(\sqrt{3}: 1\)
4 \(1: 1\) K CET-2011]
Explanation:
B Given, \(\theta_{1}=30^{\circ} \quad \theta_{2}=45^{\circ}\) We know that, Velocity \(=\) slope of \(x-t\) graph \(\mathrm{v}=\tan \theta\) \(\therefore \quad \mathrm{v}_{1}=\tan 30^{\circ}\) \(\mathrm{v}_{1}=\frac{1}{\sqrt{3}}\) \(\mathrm{v}_{2}=\tan 45^{\circ}\) \(\mathrm{v}_{2}=1\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\frac{1}{\sqrt{3}}}{1}\) \(\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{3}\)
Motion in One Dimensions
141359
The ratio of the distances travelled by a freely falling body in the \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and 4 the second
1 \(1: 3: 5: 7\)
2 \(1: 1: 1: 1\)
3 \(1: 2: 3: 4\)
4 \(1: 4: 9: 16\)
Explanation:
A Since initial velocity of freely falling body is zero. Then distance travelled by the body - \(\mathrm{s} =\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s} =0+\frac{1}{2} \mathrm{gt}^{2} \quad[\because \mathrm{g}=\text { constant }]\) \(\therefore \quad \mathrm{s} \propto \mathrm{t}^{2}\) Now distance travelled by the body in \(1 \mathrm{~s}\) \(\mathrm{s}_{1}=\mathrm{x}_{1}=\mathrm{kt}^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}(1)^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}\) Distance travelled by body in 2 sec. \(\mathrm{s}_{2}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\mathrm{k}(2)^{2}-\mathrm{k}(1)^{2}\) \(\mathrm{~s}_{2}=4 \mathrm{k}-\mathrm{k}=3 \mathrm{k}\) And just like above process- \(\mathrm{s}_{3}=\left(\mathrm{x}_{3}-\mathrm{x}_{2}\right)=\mathrm{k}(3)^{2}-\mathrm{k}(2)^{2}\) \(\mathrm{~s}_{3}=9 \mathrm{k}-4 \mathrm{k}=5 \mathrm{k}\) \(\mathrm{s}_{4}=\left(\mathrm{x}_{4}-\mathrm{x}_{3}\right)=\mathrm{k}(4)^{2}-\mathrm{k}(3)^{2}\) \(\mathrm{~s}_{4}=16 \mathrm{k}-9 \mathrm{k}=7 \mathrm{k}\) From equation (i), (ii), (iii) and (iv) \(\mathrm{s}_{1}: \mathrm{s}_{2}: \mathrm{s}_{3}: \mathrm{s}_{4}=1: 3: 5: 7\)
NEET-2022
Motion in One Dimensions
141360
The weight of an object is \(90 \mathrm{~kg}\) at the surface of the earth. If it is taken to a height equal to half of the radius of the earth, then its weight will become:
1 \(135 \mathrm{~kg}\)
2 \(45 \mathrm{~kg}\)
3 \(60 \mathrm{~kg}\)
4 \(40 \mathrm{~kg}\)
Explanation:
D We know \(\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{(1+\mathrm{h} / \mathrm{r})^{2}}\) Now, \(h=R / 2\) \(g_{h}=\frac{g}{(1+R / 2 R)^{2}}\) \(\mathrm{g}_{\mathrm{h}}=\frac{4}{9} \mathrm{~g}\) Multiplying with mass both sides. \(\mathrm{mg}_{\mathrm{h}}=\frac{4}{9} \mathrm{mg}\) \(\frac{4}{9} \times 90=40 \mathrm{~kg}\)
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Motion in One Dimensions
141357
Two cars \(P\) and \(Q\) start from a point at the same time in a straight line and their positions are represented by \(X_{P}(t)=a t+b t^{2}\) and \(X_{Q}(t)=\) \(\mathrm{ft}-\mathrm{t}^{2}\). At what time do the cars have the same velocity?
D Given: \(X_{P}(t)=a t+b t^{2} \ldots\) (i) \(\text { And } X_{Q}(t)=f t-t^{2}\) Differentiating eqn (i) and (ii) w.r.t ' \(t\) ' we get, \(\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{P}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{at}+\mathrm{bt}^{2}\right)}{\mathrm{dt}}=\mathrm{a}+2 \mathrm{bt}\) \(\mathrm{v}_{\mathrm{Q}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{Q}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{ft}-\mathrm{t}^{2}\right)}{\mathrm{dt}}=\mathrm{f}-2 \mathrm{t}\) Let at time ' \(\mathrm{t}_{0}\) ' the cars have same velocity \({\left[\mathrm{v}_{\mathrm{P}}\right]_{\mathrm{t}=\mathrm{t}_{0}}=\left[\mathrm{v}_{\mathrm{Q}}\right]_{\mathrm{t}=\mathrm{t}_{0}}}\) \(\Rightarrow \mathrm{a}+2 \mathrm{bt}_{0}=\mathrm{f}-2 \mathrm{t}_{0}\) \(\Rightarrow \mathrm{t}_{0}=\frac{\mathrm{f}-\mathrm{a}}{2(1+\mathrm{b})}\)
NEET 2016
Motion in One Dimensions
141358
The displacement-time graphs of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(x\) axis as shown in the figure. The ratio of their respective velocity is
1 \(1: 2\)
2 \(1: \sqrt{3}\)
3 \(\sqrt{3}: 1\)
4 \(1: 1\) K CET-2011]
Explanation:
B Given, \(\theta_{1}=30^{\circ} \quad \theta_{2}=45^{\circ}\) We know that, Velocity \(=\) slope of \(x-t\) graph \(\mathrm{v}=\tan \theta\) \(\therefore \quad \mathrm{v}_{1}=\tan 30^{\circ}\) \(\mathrm{v}_{1}=\frac{1}{\sqrt{3}}\) \(\mathrm{v}_{2}=\tan 45^{\circ}\) \(\mathrm{v}_{2}=1\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\frac{1}{\sqrt{3}}}{1}\) \(\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{3}\)
Motion in One Dimensions
141359
The ratio of the distances travelled by a freely falling body in the \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and 4 the second
1 \(1: 3: 5: 7\)
2 \(1: 1: 1: 1\)
3 \(1: 2: 3: 4\)
4 \(1: 4: 9: 16\)
Explanation:
A Since initial velocity of freely falling body is zero. Then distance travelled by the body - \(\mathrm{s} =\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s} =0+\frac{1}{2} \mathrm{gt}^{2} \quad[\because \mathrm{g}=\text { constant }]\) \(\therefore \quad \mathrm{s} \propto \mathrm{t}^{2}\) Now distance travelled by the body in \(1 \mathrm{~s}\) \(\mathrm{s}_{1}=\mathrm{x}_{1}=\mathrm{kt}^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}(1)^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}\) Distance travelled by body in 2 sec. \(\mathrm{s}_{2}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\mathrm{k}(2)^{2}-\mathrm{k}(1)^{2}\) \(\mathrm{~s}_{2}=4 \mathrm{k}-\mathrm{k}=3 \mathrm{k}\) And just like above process- \(\mathrm{s}_{3}=\left(\mathrm{x}_{3}-\mathrm{x}_{2}\right)=\mathrm{k}(3)^{2}-\mathrm{k}(2)^{2}\) \(\mathrm{~s}_{3}=9 \mathrm{k}-4 \mathrm{k}=5 \mathrm{k}\) \(\mathrm{s}_{4}=\left(\mathrm{x}_{4}-\mathrm{x}_{3}\right)=\mathrm{k}(4)^{2}-\mathrm{k}(3)^{2}\) \(\mathrm{~s}_{4}=16 \mathrm{k}-9 \mathrm{k}=7 \mathrm{k}\) From equation (i), (ii), (iii) and (iv) \(\mathrm{s}_{1}: \mathrm{s}_{2}: \mathrm{s}_{3}: \mathrm{s}_{4}=1: 3: 5: 7\)
NEET-2022
Motion in One Dimensions
141360
The weight of an object is \(90 \mathrm{~kg}\) at the surface of the earth. If it is taken to a height equal to half of the radius of the earth, then its weight will become:
1 \(135 \mathrm{~kg}\)
2 \(45 \mathrm{~kg}\)
3 \(60 \mathrm{~kg}\)
4 \(40 \mathrm{~kg}\)
Explanation:
D We know \(\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{(1+\mathrm{h} / \mathrm{r})^{2}}\) Now, \(h=R / 2\) \(g_{h}=\frac{g}{(1+R / 2 R)^{2}}\) \(\mathrm{g}_{\mathrm{h}}=\frac{4}{9} \mathrm{~g}\) Multiplying with mass both sides. \(\mathrm{mg}_{\mathrm{h}}=\frac{4}{9} \mathrm{mg}\) \(\frac{4}{9} \times 90=40 \mathrm{~kg}\)
141357
Two cars \(P\) and \(Q\) start from a point at the same time in a straight line and their positions are represented by \(X_{P}(t)=a t+b t^{2}\) and \(X_{Q}(t)=\) \(\mathrm{ft}-\mathrm{t}^{2}\). At what time do the cars have the same velocity?
D Given: \(X_{P}(t)=a t+b t^{2} \ldots\) (i) \(\text { And } X_{Q}(t)=f t-t^{2}\) Differentiating eqn (i) and (ii) w.r.t ' \(t\) ' we get, \(\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{P}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{at}+\mathrm{bt}^{2}\right)}{\mathrm{dt}}=\mathrm{a}+2 \mathrm{bt}\) \(\mathrm{v}_{\mathrm{Q}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{Q}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{ft}-\mathrm{t}^{2}\right)}{\mathrm{dt}}=\mathrm{f}-2 \mathrm{t}\) Let at time ' \(\mathrm{t}_{0}\) ' the cars have same velocity \({\left[\mathrm{v}_{\mathrm{P}}\right]_{\mathrm{t}=\mathrm{t}_{0}}=\left[\mathrm{v}_{\mathrm{Q}}\right]_{\mathrm{t}=\mathrm{t}_{0}}}\) \(\Rightarrow \mathrm{a}+2 \mathrm{bt}_{0}=\mathrm{f}-2 \mathrm{t}_{0}\) \(\Rightarrow \mathrm{t}_{0}=\frac{\mathrm{f}-\mathrm{a}}{2(1+\mathrm{b})}\)
NEET 2016
Motion in One Dimensions
141358
The displacement-time graphs of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(x\) axis as shown in the figure. The ratio of their respective velocity is
1 \(1: 2\)
2 \(1: \sqrt{3}\)
3 \(\sqrt{3}: 1\)
4 \(1: 1\) K CET-2011]
Explanation:
B Given, \(\theta_{1}=30^{\circ} \quad \theta_{2}=45^{\circ}\) We know that, Velocity \(=\) slope of \(x-t\) graph \(\mathrm{v}=\tan \theta\) \(\therefore \quad \mathrm{v}_{1}=\tan 30^{\circ}\) \(\mathrm{v}_{1}=\frac{1}{\sqrt{3}}\) \(\mathrm{v}_{2}=\tan 45^{\circ}\) \(\mathrm{v}_{2}=1\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\frac{1}{\sqrt{3}}}{1}\) \(\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{3}\)
Motion in One Dimensions
141359
The ratio of the distances travelled by a freely falling body in the \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and 4 the second
1 \(1: 3: 5: 7\)
2 \(1: 1: 1: 1\)
3 \(1: 2: 3: 4\)
4 \(1: 4: 9: 16\)
Explanation:
A Since initial velocity of freely falling body is zero. Then distance travelled by the body - \(\mathrm{s} =\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s} =0+\frac{1}{2} \mathrm{gt}^{2} \quad[\because \mathrm{g}=\text { constant }]\) \(\therefore \quad \mathrm{s} \propto \mathrm{t}^{2}\) Now distance travelled by the body in \(1 \mathrm{~s}\) \(\mathrm{s}_{1}=\mathrm{x}_{1}=\mathrm{kt}^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}(1)^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}\) Distance travelled by body in 2 sec. \(\mathrm{s}_{2}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\mathrm{k}(2)^{2}-\mathrm{k}(1)^{2}\) \(\mathrm{~s}_{2}=4 \mathrm{k}-\mathrm{k}=3 \mathrm{k}\) And just like above process- \(\mathrm{s}_{3}=\left(\mathrm{x}_{3}-\mathrm{x}_{2}\right)=\mathrm{k}(3)^{2}-\mathrm{k}(2)^{2}\) \(\mathrm{~s}_{3}=9 \mathrm{k}-4 \mathrm{k}=5 \mathrm{k}\) \(\mathrm{s}_{4}=\left(\mathrm{x}_{4}-\mathrm{x}_{3}\right)=\mathrm{k}(4)^{2}-\mathrm{k}(3)^{2}\) \(\mathrm{~s}_{4}=16 \mathrm{k}-9 \mathrm{k}=7 \mathrm{k}\) From equation (i), (ii), (iii) and (iv) \(\mathrm{s}_{1}: \mathrm{s}_{2}: \mathrm{s}_{3}: \mathrm{s}_{4}=1: 3: 5: 7\)
NEET-2022
Motion in One Dimensions
141360
The weight of an object is \(90 \mathrm{~kg}\) at the surface of the earth. If it is taken to a height equal to half of the radius of the earth, then its weight will become:
1 \(135 \mathrm{~kg}\)
2 \(45 \mathrm{~kg}\)
3 \(60 \mathrm{~kg}\)
4 \(40 \mathrm{~kg}\)
Explanation:
D We know \(\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{(1+\mathrm{h} / \mathrm{r})^{2}}\) Now, \(h=R / 2\) \(g_{h}=\frac{g}{(1+R / 2 R)^{2}}\) \(\mathrm{g}_{\mathrm{h}}=\frac{4}{9} \mathrm{~g}\) Multiplying with mass both sides. \(\mathrm{mg}_{\mathrm{h}}=\frac{4}{9} \mathrm{mg}\) \(\frac{4}{9} \times 90=40 \mathrm{~kg}\)
141357
Two cars \(P\) and \(Q\) start from a point at the same time in a straight line and their positions are represented by \(X_{P}(t)=a t+b t^{2}\) and \(X_{Q}(t)=\) \(\mathrm{ft}-\mathrm{t}^{2}\). At what time do the cars have the same velocity?
D Given: \(X_{P}(t)=a t+b t^{2} \ldots\) (i) \(\text { And } X_{Q}(t)=f t-t^{2}\) Differentiating eqn (i) and (ii) w.r.t ' \(t\) ' we get, \(\mathrm{v}_{\mathrm{P}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{P}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{at}+\mathrm{bt}^{2}\right)}{\mathrm{dt}}=\mathrm{a}+2 \mathrm{bt}\) \(\mathrm{v}_{\mathrm{Q}}=\frac{\mathrm{d}\left[\mathrm{X}_{\mathrm{Q}}(\mathrm{t})\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left(\mathrm{ft}-\mathrm{t}^{2}\right)}{\mathrm{dt}}=\mathrm{f}-2 \mathrm{t}\) Let at time ' \(\mathrm{t}_{0}\) ' the cars have same velocity \({\left[\mathrm{v}_{\mathrm{P}}\right]_{\mathrm{t}=\mathrm{t}_{0}}=\left[\mathrm{v}_{\mathrm{Q}}\right]_{\mathrm{t}=\mathrm{t}_{0}}}\) \(\Rightarrow \mathrm{a}+2 \mathrm{bt}_{0}=\mathrm{f}-2 \mathrm{t}_{0}\) \(\Rightarrow \mathrm{t}_{0}=\frac{\mathrm{f}-\mathrm{a}}{2(1+\mathrm{b})}\)
NEET 2016
Motion in One Dimensions
141358
The displacement-time graphs of two moving particles make angles of \(30^{\circ}\) and \(45^{\circ}\) with the \(x\) axis as shown in the figure. The ratio of their respective velocity is
1 \(1: 2\)
2 \(1: \sqrt{3}\)
3 \(\sqrt{3}: 1\)
4 \(1: 1\) K CET-2011]
Explanation:
B Given, \(\theta_{1}=30^{\circ} \quad \theta_{2}=45^{\circ}\) We know that, Velocity \(=\) slope of \(x-t\) graph \(\mathrm{v}=\tan \theta\) \(\therefore \quad \mathrm{v}_{1}=\tan 30^{\circ}\) \(\mathrm{v}_{1}=\frac{1}{\sqrt{3}}\) \(\mathrm{v}_{2}=\tan 45^{\circ}\) \(\mathrm{v}_{2}=1\) \(\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{\frac{1}{\sqrt{3}}}{1}\) \(\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{3}\)
Motion in One Dimensions
141359
The ratio of the distances travelled by a freely falling body in the \(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {rd }}\) and 4 the second
1 \(1: 3: 5: 7\)
2 \(1: 1: 1: 1\)
3 \(1: 2: 3: 4\)
4 \(1: 4: 9: 16\)
Explanation:
A Since initial velocity of freely falling body is zero. Then distance travelled by the body - \(\mathrm{s} =\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s} =0+\frac{1}{2} \mathrm{gt}^{2} \quad[\because \mathrm{g}=\text { constant }]\) \(\therefore \quad \mathrm{s} \propto \mathrm{t}^{2}\) Now distance travelled by the body in \(1 \mathrm{~s}\) \(\mathrm{s}_{1}=\mathrm{x}_{1}=\mathrm{kt}^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}(1)^{2} \Rightarrow \mathrm{s}_{1}=\mathrm{k}\) Distance travelled by body in 2 sec. \(\mathrm{s}_{2}=\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)=\mathrm{k}(2)^{2}-\mathrm{k}(1)^{2}\) \(\mathrm{~s}_{2}=4 \mathrm{k}-\mathrm{k}=3 \mathrm{k}\) And just like above process- \(\mathrm{s}_{3}=\left(\mathrm{x}_{3}-\mathrm{x}_{2}\right)=\mathrm{k}(3)^{2}-\mathrm{k}(2)^{2}\) \(\mathrm{~s}_{3}=9 \mathrm{k}-4 \mathrm{k}=5 \mathrm{k}\) \(\mathrm{s}_{4}=\left(\mathrm{x}_{4}-\mathrm{x}_{3}\right)=\mathrm{k}(4)^{2}-\mathrm{k}(3)^{2}\) \(\mathrm{~s}_{4}=16 \mathrm{k}-9 \mathrm{k}=7 \mathrm{k}\) From equation (i), (ii), (iii) and (iv) \(\mathrm{s}_{1}: \mathrm{s}_{2}: \mathrm{s}_{3}: \mathrm{s}_{4}=1: 3: 5: 7\)
NEET-2022
Motion in One Dimensions
141360
The weight of an object is \(90 \mathrm{~kg}\) at the surface of the earth. If it is taken to a height equal to half of the radius of the earth, then its weight will become:
1 \(135 \mathrm{~kg}\)
2 \(45 \mathrm{~kg}\)
3 \(60 \mathrm{~kg}\)
4 \(40 \mathrm{~kg}\)
Explanation:
D We know \(\mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{(1+\mathrm{h} / \mathrm{r})^{2}}\) Now, \(h=R / 2\) \(g_{h}=\frac{g}{(1+R / 2 R)^{2}}\) \(\mathrm{g}_{\mathrm{h}}=\frac{4}{9} \mathrm{~g}\) Multiplying with mass both sides. \(\mathrm{mg}_{\mathrm{h}}=\frac{4}{9} \mathrm{mg}\) \(\frac{4}{9} \times 90=40 \mathrm{~kg}\)
