141297
A wheel of circumference \(C\) is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of initial point of contact will be
1 \(C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
2 \(\frac{C}{2}\)
3 \(\pi \sqrt{\mathrm{C}^{2}+4}\)
4 \(C \sqrt{\frac{1}{\pi}+\frac{1}{2}}\)
Explanation:
A Let horizontal distance \(=\mathrm{x}\) and vertical distance \(=y\) Circumference of wheel \(=\mathrm{C}\) \(\therefore \quad \mathrm{x}=\frac{\mathrm{C}}{2}\) and \(\mathrm{Y}=\frac{\mathrm{C}}{\pi}\) \(\therefore\) Total displacement, \(\mathrm{d}=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}\) \(d =\sqrt{\left(\frac{C}{2}\right)^{2}+\left(\frac{C}{\pi}\right)^{2}}\) \(\therefore \quad d =C \sqrt{\frac{1}{4}+\frac{1}{\pi^{2}}}=C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
AP EAMCET(Medical)-2016
Motion in One Dimensions
141298
The distances travelled by a body starting from rest and travelling with uniform acceleration, in successive intervals of time of equal duration will be in the ratio:
1 \(1: 2: 3\)
2 \(1: 2: 4\)
3 \(1: 3: 5\)
4 \(1: 5: 9\)
Explanation:
C Distance travelled in first time interval, \(\mathrm{D}_{1}=\frac{1}{2} \cdot \mathrm{a}(\Delta \mathrm{t})^{2}\) And distance travelled in \(2^{\text {nd }}\) time interval, \(\mathrm{D}_{2}=\frac{1}{2} \cdot \mathrm{a} \cdot(2 \Delta \mathrm{t})^{2}-\frac{1}{2} \cdot \mathrm{a} \cdot(\Delta \mathrm{t})^{2}\) \(\mathrm{D}_{2}=\frac{3}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) Similarly, distance travelled in \(3^{\text {rd }}\) time interval, \(D_{3}=\frac{1}{2} \cdot a(3 \Delta t)^{2}-\frac{1}{2} \cdot a(2 \Delta t)^{2}\) \(\mathrm{D}_{3}=\frac{5}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) So, \(\mathrm{D}_{1}: \mathrm{D}_{2}: \mathrm{D}_{3}=1: 3: 5\)
AP EAMCET(Medical)-1999
Motion in One Dimensions
141299
A stone is thrown vertically upwards with a velocity \(30 \mathrm{~ms}^{-1}\). If the acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\), what is the distance travelled by the particle during the first second of its motion?
1 \(10 \mathrm{~m}\)
2 \(25 \mathrm{~m}\)
3 \(30 \mathrm{~m}\)
4 None of the above
Explanation:
B Given: Initial velocity, \(u=30 \mathrm{~m} / \mathrm{s}\) Distance travelled during the first second of motion will be \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\mathrm{~s}=30 \times 1-\frac{1}{2} \mathrm{~g}(1)^{2} \quad[\because \mathrm{a}=-\mathrm{g}]\) \(\mathrm{s}=30 \times 1-\frac{1}{2} \times 10\) \(\mathrm{~s}=25 \mathrm{~m}\)
SRM JEE - 2009
Motion in One Dimensions
141300
What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey?
1 \(4: 5\)
2 \(7: 9\)
3 \(16: 25\)
4 \(1: 1\)
Explanation:
B Distance moved by a freely falling body from rest in 4 second is given by \(\mathrm{s}_{\mathrm{n}}=\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}_{4}=\frac{1}{2} \mathrm{~g}(4)^{2}=\frac{1}{2} \mathrm{~g} \times 16\) Similarly \(\mathrm{s}_{5}=\frac{1}{2} \mathrm{~g}(5)^{2}=\frac{1}{2} \mathrm{~g} \times 25\) \(\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g}(3)^{2}=\frac{1}{2} \mathrm{~g} \times 9\) Distance moved in \(5^{\text {th }}\) second \(=(\) distance moved in 5 second)-(distance moved in 4 second) \(=\mathrm{s}_{5}-\mathrm{s}_{4}=\frac{1}{2} \mathrm{~g} \times 25-\frac{1}{2} \mathrm{~g} \times 16\) \(=\frac{1}{2} \mathrm{~g}(25-16)\) Distance moved in \(4^{\text {th }}\) second - \(\mathrm{s}_{4}-\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g} \times 16-\frac{1}{2} \mathrm{~g} \times 9\) \(=\frac{1}{2} \mathrm{~g}(16-9)\) Ratio of distance moved in \(4^{\text {th }}\) and \(5^{\text {th }}\) second \(=\frac{\frac{1}{2} g(16-9)}{\frac{1}{2} g(25-16)}\) \(=\frac{16-9}{25-16}=\frac{7}{9}\)
141297
A wheel of circumference \(C\) is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of initial point of contact will be
1 \(C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
2 \(\frac{C}{2}\)
3 \(\pi \sqrt{\mathrm{C}^{2}+4}\)
4 \(C \sqrt{\frac{1}{\pi}+\frac{1}{2}}\)
Explanation:
A Let horizontal distance \(=\mathrm{x}\) and vertical distance \(=y\) Circumference of wheel \(=\mathrm{C}\) \(\therefore \quad \mathrm{x}=\frac{\mathrm{C}}{2}\) and \(\mathrm{Y}=\frac{\mathrm{C}}{\pi}\) \(\therefore\) Total displacement, \(\mathrm{d}=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}\) \(d =\sqrt{\left(\frac{C}{2}\right)^{2}+\left(\frac{C}{\pi}\right)^{2}}\) \(\therefore \quad d =C \sqrt{\frac{1}{4}+\frac{1}{\pi^{2}}}=C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
AP EAMCET(Medical)-2016
Motion in One Dimensions
141298
The distances travelled by a body starting from rest and travelling with uniform acceleration, in successive intervals of time of equal duration will be in the ratio:
1 \(1: 2: 3\)
2 \(1: 2: 4\)
3 \(1: 3: 5\)
4 \(1: 5: 9\)
Explanation:
C Distance travelled in first time interval, \(\mathrm{D}_{1}=\frac{1}{2} \cdot \mathrm{a}(\Delta \mathrm{t})^{2}\) And distance travelled in \(2^{\text {nd }}\) time interval, \(\mathrm{D}_{2}=\frac{1}{2} \cdot \mathrm{a} \cdot(2 \Delta \mathrm{t})^{2}-\frac{1}{2} \cdot \mathrm{a} \cdot(\Delta \mathrm{t})^{2}\) \(\mathrm{D}_{2}=\frac{3}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) Similarly, distance travelled in \(3^{\text {rd }}\) time interval, \(D_{3}=\frac{1}{2} \cdot a(3 \Delta t)^{2}-\frac{1}{2} \cdot a(2 \Delta t)^{2}\) \(\mathrm{D}_{3}=\frac{5}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) So, \(\mathrm{D}_{1}: \mathrm{D}_{2}: \mathrm{D}_{3}=1: 3: 5\)
AP EAMCET(Medical)-1999
Motion in One Dimensions
141299
A stone is thrown vertically upwards with a velocity \(30 \mathrm{~ms}^{-1}\). If the acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\), what is the distance travelled by the particle during the first second of its motion?
1 \(10 \mathrm{~m}\)
2 \(25 \mathrm{~m}\)
3 \(30 \mathrm{~m}\)
4 None of the above
Explanation:
B Given: Initial velocity, \(u=30 \mathrm{~m} / \mathrm{s}\) Distance travelled during the first second of motion will be \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\mathrm{~s}=30 \times 1-\frac{1}{2} \mathrm{~g}(1)^{2} \quad[\because \mathrm{a}=-\mathrm{g}]\) \(\mathrm{s}=30 \times 1-\frac{1}{2} \times 10\) \(\mathrm{~s}=25 \mathrm{~m}\)
SRM JEE - 2009
Motion in One Dimensions
141300
What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey?
1 \(4: 5\)
2 \(7: 9\)
3 \(16: 25\)
4 \(1: 1\)
Explanation:
B Distance moved by a freely falling body from rest in 4 second is given by \(\mathrm{s}_{\mathrm{n}}=\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}_{4}=\frac{1}{2} \mathrm{~g}(4)^{2}=\frac{1}{2} \mathrm{~g} \times 16\) Similarly \(\mathrm{s}_{5}=\frac{1}{2} \mathrm{~g}(5)^{2}=\frac{1}{2} \mathrm{~g} \times 25\) \(\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g}(3)^{2}=\frac{1}{2} \mathrm{~g} \times 9\) Distance moved in \(5^{\text {th }}\) second \(=(\) distance moved in 5 second)-(distance moved in 4 second) \(=\mathrm{s}_{5}-\mathrm{s}_{4}=\frac{1}{2} \mathrm{~g} \times 25-\frac{1}{2} \mathrm{~g} \times 16\) \(=\frac{1}{2} \mathrm{~g}(25-16)\) Distance moved in \(4^{\text {th }}\) second - \(\mathrm{s}_{4}-\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g} \times 16-\frac{1}{2} \mathrm{~g} \times 9\) \(=\frac{1}{2} \mathrm{~g}(16-9)\) Ratio of distance moved in \(4^{\text {th }}\) and \(5^{\text {th }}\) second \(=\frac{\frac{1}{2} g(16-9)}{\frac{1}{2} g(25-16)}\) \(=\frac{16-9}{25-16}=\frac{7}{9}\)
141297
A wheel of circumference \(C\) is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of initial point of contact will be
1 \(C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
2 \(\frac{C}{2}\)
3 \(\pi \sqrt{\mathrm{C}^{2}+4}\)
4 \(C \sqrt{\frac{1}{\pi}+\frac{1}{2}}\)
Explanation:
A Let horizontal distance \(=\mathrm{x}\) and vertical distance \(=y\) Circumference of wheel \(=\mathrm{C}\) \(\therefore \quad \mathrm{x}=\frac{\mathrm{C}}{2}\) and \(\mathrm{Y}=\frac{\mathrm{C}}{\pi}\) \(\therefore\) Total displacement, \(\mathrm{d}=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}\) \(d =\sqrt{\left(\frac{C}{2}\right)^{2}+\left(\frac{C}{\pi}\right)^{2}}\) \(\therefore \quad d =C \sqrt{\frac{1}{4}+\frac{1}{\pi^{2}}}=C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
AP EAMCET(Medical)-2016
Motion in One Dimensions
141298
The distances travelled by a body starting from rest and travelling with uniform acceleration, in successive intervals of time of equal duration will be in the ratio:
1 \(1: 2: 3\)
2 \(1: 2: 4\)
3 \(1: 3: 5\)
4 \(1: 5: 9\)
Explanation:
C Distance travelled in first time interval, \(\mathrm{D}_{1}=\frac{1}{2} \cdot \mathrm{a}(\Delta \mathrm{t})^{2}\) And distance travelled in \(2^{\text {nd }}\) time interval, \(\mathrm{D}_{2}=\frac{1}{2} \cdot \mathrm{a} \cdot(2 \Delta \mathrm{t})^{2}-\frac{1}{2} \cdot \mathrm{a} \cdot(\Delta \mathrm{t})^{2}\) \(\mathrm{D}_{2}=\frac{3}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) Similarly, distance travelled in \(3^{\text {rd }}\) time interval, \(D_{3}=\frac{1}{2} \cdot a(3 \Delta t)^{2}-\frac{1}{2} \cdot a(2 \Delta t)^{2}\) \(\mathrm{D}_{3}=\frac{5}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) So, \(\mathrm{D}_{1}: \mathrm{D}_{2}: \mathrm{D}_{3}=1: 3: 5\)
AP EAMCET(Medical)-1999
Motion in One Dimensions
141299
A stone is thrown vertically upwards with a velocity \(30 \mathrm{~ms}^{-1}\). If the acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\), what is the distance travelled by the particle during the first second of its motion?
1 \(10 \mathrm{~m}\)
2 \(25 \mathrm{~m}\)
3 \(30 \mathrm{~m}\)
4 None of the above
Explanation:
B Given: Initial velocity, \(u=30 \mathrm{~m} / \mathrm{s}\) Distance travelled during the first second of motion will be \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\mathrm{~s}=30 \times 1-\frac{1}{2} \mathrm{~g}(1)^{2} \quad[\because \mathrm{a}=-\mathrm{g}]\) \(\mathrm{s}=30 \times 1-\frac{1}{2} \times 10\) \(\mathrm{~s}=25 \mathrm{~m}\)
SRM JEE - 2009
Motion in One Dimensions
141300
What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey?
1 \(4: 5\)
2 \(7: 9\)
3 \(16: 25\)
4 \(1: 1\)
Explanation:
B Distance moved by a freely falling body from rest in 4 second is given by \(\mathrm{s}_{\mathrm{n}}=\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}_{4}=\frac{1}{2} \mathrm{~g}(4)^{2}=\frac{1}{2} \mathrm{~g} \times 16\) Similarly \(\mathrm{s}_{5}=\frac{1}{2} \mathrm{~g}(5)^{2}=\frac{1}{2} \mathrm{~g} \times 25\) \(\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g}(3)^{2}=\frac{1}{2} \mathrm{~g} \times 9\) Distance moved in \(5^{\text {th }}\) second \(=(\) distance moved in 5 second)-(distance moved in 4 second) \(=\mathrm{s}_{5}-\mathrm{s}_{4}=\frac{1}{2} \mathrm{~g} \times 25-\frac{1}{2} \mathrm{~g} \times 16\) \(=\frac{1}{2} \mathrm{~g}(25-16)\) Distance moved in \(4^{\text {th }}\) second - \(\mathrm{s}_{4}-\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g} \times 16-\frac{1}{2} \mathrm{~g} \times 9\) \(=\frac{1}{2} \mathrm{~g}(16-9)\) Ratio of distance moved in \(4^{\text {th }}\) and \(5^{\text {th }}\) second \(=\frac{\frac{1}{2} g(16-9)}{\frac{1}{2} g(25-16)}\) \(=\frac{16-9}{25-16}=\frac{7}{9}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141297
A wheel of circumference \(C\) is at rest on the ground. When the wheel rolls forward through half a revolution, then the displacement of initial point of contact will be
1 \(C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
2 \(\frac{C}{2}\)
3 \(\pi \sqrt{\mathrm{C}^{2}+4}\)
4 \(C \sqrt{\frac{1}{\pi}+\frac{1}{2}}\)
Explanation:
A Let horizontal distance \(=\mathrm{x}\) and vertical distance \(=y\) Circumference of wheel \(=\mathrm{C}\) \(\therefore \quad \mathrm{x}=\frac{\mathrm{C}}{2}\) and \(\mathrm{Y}=\frac{\mathrm{C}}{\pi}\) \(\therefore\) Total displacement, \(\mathrm{d}=\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}}\) \(d =\sqrt{\left(\frac{C}{2}\right)^{2}+\left(\frac{C}{\pi}\right)^{2}}\) \(\therefore \quad d =C \sqrt{\frac{1}{4}+\frac{1}{\pi^{2}}}=C \sqrt{\frac{1}{\pi^{2}}+\frac{1}{4}}\)
AP EAMCET(Medical)-2016
Motion in One Dimensions
141298
The distances travelled by a body starting from rest and travelling with uniform acceleration, in successive intervals of time of equal duration will be in the ratio:
1 \(1: 2: 3\)
2 \(1: 2: 4\)
3 \(1: 3: 5\)
4 \(1: 5: 9\)
Explanation:
C Distance travelled in first time interval, \(\mathrm{D}_{1}=\frac{1}{2} \cdot \mathrm{a}(\Delta \mathrm{t})^{2}\) And distance travelled in \(2^{\text {nd }}\) time interval, \(\mathrm{D}_{2}=\frac{1}{2} \cdot \mathrm{a} \cdot(2 \Delta \mathrm{t})^{2}-\frac{1}{2} \cdot \mathrm{a} \cdot(\Delta \mathrm{t})^{2}\) \(\mathrm{D}_{2}=\frac{3}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) Similarly, distance travelled in \(3^{\text {rd }}\) time interval, \(D_{3}=\frac{1}{2} \cdot a(3 \Delta t)^{2}-\frac{1}{2} \cdot a(2 \Delta t)^{2}\) \(\mathrm{D}_{3}=\frac{5}{2} \cdot \mathrm{a} \Delta \mathrm{t}^{2}\) So, \(\mathrm{D}_{1}: \mathrm{D}_{2}: \mathrm{D}_{3}=1: 3: 5\)
AP EAMCET(Medical)-1999
Motion in One Dimensions
141299
A stone is thrown vertically upwards with a velocity \(30 \mathrm{~ms}^{-1}\). If the acceleration due to gravity is \(10 \mathrm{~ms}^{-2}\), what is the distance travelled by the particle during the first second of its motion?
1 \(10 \mathrm{~m}\)
2 \(25 \mathrm{~m}\)
3 \(30 \mathrm{~m}\)
4 None of the above
Explanation:
B Given: Initial velocity, \(u=30 \mathrm{~m} / \mathrm{s}\) Distance travelled during the first second of motion will be \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\mathrm{~s}=30 \times 1-\frac{1}{2} \mathrm{~g}(1)^{2} \quad[\because \mathrm{a}=-\mathrm{g}]\) \(\mathrm{s}=30 \times 1-\frac{1}{2} \times 10\) \(\mathrm{~s}=25 \mathrm{~m}\)
SRM JEE - 2009
Motion in One Dimensions
141300
What will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey?
1 \(4: 5\)
2 \(7: 9\)
3 \(16: 25\)
4 \(1: 1\)
Explanation:
B Distance moved by a freely falling body from rest in 4 second is given by \(\mathrm{s}_{\mathrm{n}}=\frac{1}{2} \mathrm{gt}^{2}\) \(\mathrm{~s}_{4}=\frac{1}{2} \mathrm{~g}(4)^{2}=\frac{1}{2} \mathrm{~g} \times 16\) Similarly \(\mathrm{s}_{5}=\frac{1}{2} \mathrm{~g}(5)^{2}=\frac{1}{2} \mathrm{~g} \times 25\) \(\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g}(3)^{2}=\frac{1}{2} \mathrm{~g} \times 9\) Distance moved in \(5^{\text {th }}\) second \(=(\) distance moved in 5 second)-(distance moved in 4 second) \(=\mathrm{s}_{5}-\mathrm{s}_{4}=\frac{1}{2} \mathrm{~g} \times 25-\frac{1}{2} \mathrm{~g} \times 16\) \(=\frac{1}{2} \mathrm{~g}(25-16)\) Distance moved in \(4^{\text {th }}\) second - \(\mathrm{s}_{4}-\mathrm{s}_{3}=\frac{1}{2} \mathrm{~g} \times 16-\frac{1}{2} \mathrm{~g} \times 9\) \(=\frac{1}{2} \mathrm{~g}(16-9)\) Ratio of distance moved in \(4^{\text {th }}\) and \(5^{\text {th }}\) second \(=\frac{\frac{1}{2} g(16-9)}{\frac{1}{2} g(25-16)}\) \(=\frac{16-9}{25-16}=\frac{7}{9}\)
