[A]: The two sperms cells in a pollen tube often change their shape.
[R]: The sperms are released in the synergid as intact cells but only their nuclei migrate
[A]: The two sperms cells in a pollen tube often change their shape.
[R]: The sperms are released in the synergid as intact cells but only their nuclei migrate
[R]: The sperms are released in the synergid as intact cells but only their nuclei migrate
A
If both A and R are true and R is the correct explanation of A
B
If both A and R are true but R is not the correct explanation of A
C
If A is true and R is false
D
If both A and R are false
Text Solution
AI Generated Solution
The correct Answer is:
To solve the question regarding the assertion and reason about sperm cells in a pollen tube, we can break it down into a step-by-step explanation.
### Step-by-Step Solution:
1. **Understanding the Assertion (A)**:
- The assertion states that "the two sperm cells in a pollen tube often change their shape."
- In a pollen tube, there are two sperm cells (male gametes) that are released during the process of pollination. These sperm cells can change shape as they move through the pollen tube.
**Hint**: Think about how cells can adapt their shape for movement and interaction with other cells.
2. **Understanding the Reason (R)**:
- The reason states that "the sperms are released in the synergid as intact cells but only their nuclei migrate."
- When the pollen tube reaches the ovule, it penetrates the synergid cells. The sperm cells are released as intact cells, but during fertilization, only the nuclei of these sperm cells migrate to fertilize the egg cell and the central cell.
**Hint**: Consider the process of fertilization and how sperm cells interact with the female reproductive structures.
3. **Evaluating the Truth of A and R**:
- Both the assertion and the reason are true. The sperm cells do change shape while moving through the pollen tube, and they are released intact in the synergid, with only their nuclei migrating for fertilization.
**Hint**: Reflect on the mechanisms of pollen tube growth and fertilization to confirm the accuracy of both statements.
4. **Determining the Relationship Between A and R**:
- While both statements are true, the reason (R) does not directly explain the assertion (A). The shape change of the sperm cells is not necessarily a result of their nuclei migrating; rather, it is a characteristic of the sperm cells themselves.
**Hint**: Analyze whether the reason provides a valid explanation for the assertion or if they are simply related facts.
5. **Conclusion**:
- The correct answer is that both A and R are true, but R is not a correct explanation of A.
**Hint**: Make sure to differentiate between statements that are true and those that provide a logical explanation for each other.
### Final Answer:
Both A and R are true, but R is not a correct explanation of A.
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Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends on the properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} Which object would first acquire half of their respective terminal speed in minimum time from start of the motion of all were released simultaneously ?
Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} At the start of motion when object is released in the liquid, its acceleration is :
Direction : Resistive force proportional to object velocity At low speeds, the resistive force acting on an object that is moving a viscous medium is effectively modeleld as being proportional to the object velocity. The mathematical representation of the resistive force can be expressed as R = -bv Where v is the velocity of the object and b is a positive constant that depends onthe properties of the medium and on the shape and dimensions of the object. The negative sign represents the fact that the resistance froce is opposite to the velocity. Consider a sphere of mass m released frm rest in a liquid. Assuming that the only forces acting on the spheres are the resistive froce R and the weight mg, we can describe its motion using Newton's second law. though the buoyant force is also acting on the submerged object the force is constant and effect of this force be modeled by changing the apparent weight of the sphere by a constant froce, so we can ignore it here. Thus mg - bv = m (dv)/(dt) rArr (dv)/(dt) = g - (b)/(m) v Solving the equation v = (mg)/(b) (1- e^(-bt//m)) where e=2.71 is the base of the natural logarithm The acceleration becomes zero when the increasing resistive force eventually the weight. At this point, the object reaches its terminals speed v_(1) and then on it continues to move with zero acceleration mg - b_(T) =0 rArr m_(T) = (mg)/(b) Hence v = v_(T) (1-e^((vt)/(m))) In an experimental set-up four objects I,II,III,IV were released in same liquid. Using the data collected for the subsequent motions value of constant b were calculated. Respective data are shown in table. {:("Object",I,II,II,IV),("Mass (in kg.)",1,2,3,4),(underset("in (N-s)/m")("Constant b"),3.7,1.4,1.4,2.8):} Which object has greatest terminal speed in the liquid ?
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